**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases.

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
import java.io.*;
import java.util.*;

public class Main {

    static int grid[][] = {{0, 0, 1, -1, 1, 1, -1, -1}, {1, -1, 0, 0, 1, -1, 1, -1}};
    static long[][] dp2;
    static int[][] dp;
    static int mod = (int) 1e9 + 7;
    static int mo = 998244353;
    static int nu;
    static ArrayList<Integer> conCom;
    static int num[], lowesLink[], max1[], max2[];
    static boolean inStack[];
    static Stack<Integer> st;
    static LinkedList<ArrayList<Integer>> stronConCom;

    static boolean cycle[];
    static boolean vi[], s[];
    static boolean ca;
    static ArrayList<Integer>[] g, rg;
    static int oo = (int) 1e9;
    static Reader input = new Reader();

    static BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
    static int anc[][], depth[], in[], out[], timer;
    static ArrayList<tp> ans;
    static long cost[];
    static ArrayList<Integer> primeFactors, nodes;
    static boolean big[];

    public static void main(String[] args) throws IOException {
        //Reader input = new Reader("shuffle.in");
        //BufferedWriter log = new BufferedWriter(new OutputStreamWriter(new FileOutputStream("shuffle.out")));
        int test = input.nextInt();
        next:
        for (int te = 1; te <= test; te++) {
            long n = input.nextInt();
            long k = input.nextInt();
            long max = n + k - 1;
            long r = n;
            long l = 1;
            long id = -1;
            long e = k * (k - 1) / 2;
            max = max * (max + 1) / 2 - e;
            while (r >= l) {
                long m = (r + l) / 2;
                long c = k + m;
                long v = c * (c + 1) / 2 - e;
                if (v <= max - v) {
                    id = m;
                    l = m + 1;
                } else {
                    r = m - 1;
                }
            }
            long c = k + id;
            long v = c * (c + 1) / 2 - e;
            long ans = (max - v * 2L);
            c = k + id + 1;
            if (c <= n + k - 1) {
                v = c * (c + 1) / 2 - e;
                ans = Math.min(ans, Math.abs(max - v * 2L));
            }
            log.write(ans + "\n");
        }
        log.flush();
    }

    static int getNumberOfOnes(long x) {
        int cnt = 0;
        while (x >= 1) {
            if ((x & 1) > 0) cnt++;
            x >>= 1;
        }
        return cnt;
    }

    static boolean isPal(int l, int r, int n, HashedString h, HashedString rh) {
        long v = h.getHash(l, r);
        long u = rh.getHash(n - r - 1, n - l - 1);
        return v == u;
    }

    static void dfs(int node, int par) {
        anc[node][0] = par;
        in[node] = timer++;
        rg[depth[node]].add(node);
        for (Integer ch : g[node]) {
            if (ch == par) continue;
            depth[ch] = depth[node] + 1;
            dfs(ch, node);
        }
        out[node] = timer - 1;
    }

    static long[] dij(int node, ArrayList<pair> g[]) {
        PriorityQueue<pai> q = new PriorityQueue<>(new Comparator<pai>() {
            @Override
            public int compare(pai o1, pai o2) {
                return Long.compare(o1.y, o2.y);
            }
        });
        q.add(new pai(node, 0));
        long distance[] = new long[g.length];
        Arrays.fill(distance, -1);
        while (!q.isEmpty()) {
            pai p = q.poll();
            long cost = p.y;
            if (distance[p.x] != -1) {
                continue;
            }
            distance[p.x] = cost;
            ArrayList<pair> nodes = g[p.x];
            for (pair node1 : nodes) {
                if (distance[node1.x] == -1) {
                    pai pa = new pai(node1.x, cost + node1.y);
                    q.add(pa);
                }
            }
        }
        return distance;
    }

    static int lca(int u, int v) {
        if (depth[u] < depth[v]) {
            int temp = u;
            u = v;
            v = temp;
        }
        u = lift(u, depth[u] - depth[v], anc);

        if (u == v) return u;
        for (int i = anc[0].length - 1; i >= 0; i--) {
            if (anc[u][i] != anc[v][i]) {
                u = anc[u][i];
                v = anc[v][i];
            }
        }
        return anc[u][0];
    }

    static int lift(int v, int d, int anc[][]) {
        int p = 1;
        for (int i = 0; i < anc[0].length; i++) {
            if ((p & d) > 0) {
                v = anc[v][i];
            }
            p <<= 1;
        }
        return v;
    }

    static class FenwickTree {
        private int[] bit;
        private int n;

        // Constructor to initialize the BIT with a given size
        public FenwickTree(int size) {
            this.n = size;
            this.bit = new int[size + 1]; // BIT is 1-indexed, so we need an array of size n+1
        }

        // Method to update the BIT at index 'index' by adding 'value'
        public void update(int index, int value) {
            int currentValue = query(index) - query(index - 1); // Find the current value at index
            int delta = value - currentValue; // Calculate the difference needed to update the value

            // Update BIT with the difference
            for (; index <= n; index += index & -index) {
                bit[index] += delta;
            }
        }

        // Method to get the prefix sum up to index 'index'
        public int query(int index) {
            int sum = 0;
            for (; index > 0; index -= index & -index) {
                sum += bit[index];
            }
            return sum;
        }

        // Method to get the sum between two indices (inclusive)
        public int rangeQuery(int left, int right) {
            return query(right) - query(left - 1);
        }

        // Method to initialize BIT with an array (1-indexed)
        public void buildFromArray(int[] arr) {
            for (int i = 1; i <= n; i++) {
                update(i, arr[i]);
            }
        }
    }

    static class SegmentTreeEdu {
        private class Node {
            int value;

            public Node(int value) {
                this.value = value;
            }
        }

        private Node seg[];
        private int size;
        private int arr[];

        public SegmentTreeEdu(int a[]) {
            seg = new Node[a.length * 4];
            size = a.length - 1;
            arr = a.clone();
            build(1, 1, size, a);
        }

        public SegmentTreeEdu(int n) {
            seg = new Node[(n + 1) * 4];
            size = n;
            arr = new int[n + 1];
            build(1, 1, size, arr);
        }

        private Node merge(Node left, Node right) {
            return new Node(left.value + right.value);
        }

        private void build(int idx, int l, int r, int a[]) {
            if (l == r) {
                seg[idx] = new Node(a[l]);
                return;
            }
            build(idx * 2, l, (l + r) / 2, a);
            build(idx * 2 + 1, (l + r) / 2 + 1, r, a);
            seg[idx] = merge(seg[idx * 2], seg[idx * 2 + 1]);
        }

        private void update(int idx, int l, int r, int index, int val) {
            if (index > r || index < l) return;
            if (index == l && index == r) {
                arr[l] = val;
                seg[idx] = new Node(arr[l]);
                return;
            }
            update(idx * 2, l, (l + r) / 2, index, val);
            update(idx * 2 + 1, (l + r) / 2 + 1, r, index, val);
            seg[idx] = merge(seg[idx * 2], seg[idx * 2 + 1]);
        }

        public void update(int index, int val) {
            update(1, 1, size, index, val);
        }

        public Node getSum(int idx, int s, int e, int l, int r) {
            if (r < s || l > e) return new Node(0);
            if (s >= l && e <= r) return new Node(seg[idx].value);
            return merge(getSum(idx * 2, s, (s + e) / 2, l, r), getSum(idx * 2 + 1, (s + e) / 2 + 1, e, l, r));
        }

        public int getAns(int l, int r) {
            return getSum(1, 1, size, l, r).value;
        }
    }

    static class tp {
        int x, y, z, w;

        public tp(int x, int y, int z, int w) {
            this.x = x;
            this.y = y;
            this.z = z;
            this.w = w;
        }
    }

    static int first(int suc[], int x) {
        int a = suc[x];
        int b = suc[a];
        while (a != b) {
            a = suc[a];
            b = suc[suc[b]];
        }
        a = x;
        while (a != b) {
            a = suc[a];
            b = suc[b];
        }
        return a;
    }

    static ArrayList<Integer> topologicalSort(ArrayList<pair> g[], int fre[]) {
        Queue<Integer> q = new ArrayDeque<>();
        for (int i = 1; i < fre.length; i++) {
            if (fre[i] == 0) q.add(i);
        }
        ArrayList<Integer> nodes = new ArrayList<>();
        while (!q.isEmpty()) {
            int node = q.poll();
            nodes.add(node);
            for (pair p : g[node]) {
                fre[p.x]--;
                if (fre[p.x] == 0) q.add(p.x);
            }
        }
        return nodes;
    }

    static long[] getMaxPath(ArrayList<pair> g[], ArrayList<Integer> nodes) {
        long[] dis = new long[g.length];
        Arrays.fill(dis, -1);
        for (Integer node : nodes) {
            for (pair p : g[node]) {
                if (dis[node] == -1) dis[node] = 0;
                dis[p.x] = Math.max(dis[p.x], dis[node] + p.y);
            }
        }
        return dis;
    }

    public static int gcd(int x, int y) {
        while (y != 0) {
            int c = x % y;
            x = y;
            y = c;
        }
        return x;
    }

    static int bs(int idx, ArrayList<Integer> a) {
        int max = a.size() - 1;
        int min = 0;
        int ans = 0;
        while (max >= min) {
            int mid = (max + min) / 2;
            if (a.get(mid) > idx) {
                ans = a.size() - mid;
                max = mid - 1;
            } else {
                min = mid + 1;
            }
        }
        return ans;
    }

    static class advancedSegmentTree {
        private ArrayList<Integer> distinctElements;
        private SegmentTree seg;
        private SegmentTree segSum;
        private int size;

        public advancedSegmentTree() {
            distinctElements = new ArrayList<>();
            for (int i = (int) -4e5 - 5; i <= 4e5 + 5; i++) {
                distinctElements.add(i);
            }
            size = distinctElements.size() + 5;
            seg = new SegmentTree(size);
            segSum = new SegmentTree(size);
        }

        public advancedSegmentTree(int a[]) {
            distinctElements = new ArrayList<>();
            Arrays.sort(a);
            distinctElements.add(a[0]);
            for (int i = 1; i < a.length; i++) {
                if (a[i] != distinctElements.get(distinctElements.size() - 1)) {
                    distinctElements.add(a[i]);
                }
            }
            size = distinctElements.size() + 5;
            seg = new SegmentTree(size);
            segSum = new SegmentTree(size);
        }

        public advancedSegmentTree(int len) {
            distinctElements = new ArrayList<>();
            for (int i = 0; i <= len + 5; i++) {
                distinctElements.add(i);
            }
            size = distinctElements.size() + 5;
            seg = new SegmentTree(size);
            segSum = new SegmentTree(size);
        }

        public void add(int val) {
            int idx = val + 2;//getIndex(val);
            seg.updateIndex(1, 1, size, idx, 1);
            segSum.updateIndex(1, 1, size, idx, val);
        }

        public void remove(int val) {
            int idx = val + 2;// getIndex(val);
            seg.updateIndex(1, 1, size, idx, -1);
            segSum.updateIndex(1, 1, size, idx, -val);
        }

        public int getSum(int val) {
            return segSum.sumInRange(1, 1, size, 1, val - 1);
        }

        private int getIndex(int val) {
            int max = distinctElements.size() - 1;
            int min = 0;
            int ans = -1;
            while (max >= min) {
                int mid = (max + min) / 2;
                if (distinctElements.get(mid) == val) {
                    return mid + 2;
                } else if (distinctElements.get(mid) > val) {
                    max = mid - 1;
                } else {
                    min = mid + 1;
                }
            }
            return ans;
        }

        public long countNumbersGreaterThan(int val) {
            int idx = getIndex(val);
            return seg.sumInRange(1, 1, size, idx + 1, size);
        }

        public long countNumbersEqualTo(int val) {
            int idx = getIndex(val);
            return seg.sumInRange(1, 1, size, idx, idx);
        }

        public long countNumbersLessThan(int val) {
            int idx = getIndex(val);
            return seg.sumInRange(1, 1, size, 1, idx - 1);
        }

        private static class SegmentTree {

            int size;
            int seg[], la[];

            public SegmentTree(int size) {
                this.size = size;
                seg = new int[size * 4];
                //  lazy = new long[seg.length];
            }

            void process(int idx, int s, int e) {
                seg[idx] += la[idx];
                if (s < e) {
                    la[idx * 2] += la[idx];
                    la[idx * 2 + 1] += la[idx];
                }
                la[idx] = 0;
            }

            void updateRange(int idx, int s, int e, int l, int r, int val) {
                process(idx, s, e);
                if ((l > e) || s > r) {
                    return;
                }
                if (s >= l && e <= r) {
                    la[idx] += val;
                    process(idx, s, e);
                    return;
                }
                updateRange(idx * 2, s, (s + e) / 2, l, r, val);
                updateRange(idx * 2 + 1, (s + e) / 2 + 1, e, l, r, val);
                seg[idx] = seg[idx * 2] + seg[idx * 2 + 1];
            }

            void updateIndex(int idx, int s, int e, int ind, int val) {
                //     process(idx, s, e);
                if (ind < s || ind > e) {
                    return;
                }
                if (s == ind && ind == e) {
                    seg[idx] += val;
                    return;
                }
                updateIndex(idx << 1, s, (s + e) / 2, ind, val);
                updateIndex(idx << 1 | 1, (s + e) / 2 + 1, e, ind, val);
                seg[idx] = seg[idx << 1] + seg[idx << 1 | 1];
            }

            int sumInRange(int idx, int s, int e, int l, int r) {
                //   process(idx, s, e);
                if ((l > e) || s > r) {
                    return 0;
                }
                if (s >= l && e <= r) {
                    return seg[idx];
                }
                return sumInRange(idx << 1, s, (s + e) / 2, l, r) + sumInRange(idx << 1 | 1, (s + e) / 2 + 1, e, l, r);
            }
        }
    }

    static class LowestCommonAncestor {

        private final int nodes[];
        private final int first[];
        private final int last[];
        private final int height[];
        static int cnt;
        private SegmentTreeMin seg;

        public LowestCommonAncestor(ArrayList<Integer> g[]) {
            nodes = new int[g.length * 2 - 1];
            first = new int[g.length];
            height = new int[g.length];
            last = new int[g.length];
            Arrays.fill(first, -1);
            cnt = 0;
            dfs(0, g, 1);
            // seg = new SegmentTreeMin(nodes.length - 1, nodes);
        }

        public LowestCommonAncestor(ArrayList<Integer>[] g, int root) {
            nodes = new int[g.length * 2 - 1];
            first = new int[g.length];
            height = new int[g.length];
            last = new int[g.length];
            Arrays.fill(first, -1);
            cnt = 0;
            dfs(root, g, 1);
            seg = new SegmentTreeMin(nodes.length - 1, nodes);
        }

        private void dfs(int node, ArrayList<Integer> g[], int lev) {
            first[node] = cnt;
            nodes[cnt++] = node;
            height[node] = lev;
            for (Integer ch : g[node]) {
                if (first[ch] == -1) {
                    dfs(ch, g, lev + 1);
                    nodes[cnt++] = node;
                }
            }
            last[node] = cnt;
        }

        public int findLCA(int node1, int node2) {
            int idx1 = first[node1];
            int idx2 = first[node2];
            if (idx1 > idx2) {
                int c = idx1;
                idx1 = idx2;
                idx2 = c;
            }
            return seg.query(1, 0, nodes.length - 1, idx1, idx2);
        }

        private class SegmentTreeMin {

            int size;
            int seg[];

            public SegmentTreeMin(int size, int euler[]) {
                this.size = size;
                seg = new int[size * 4];
                build(1, 0, size, euler);
            }

            private void build(int node, int b, int e, int euler[]) {
                if (b == e) {
                    seg[node] = euler[b];
                } else {
                    int mid = (b + e) / 2;
                    build(node << 1, b, mid, euler);
                    build(node << 1 | 1, mid + 1, e, euler);
                    int l = seg[node << 1], r = seg[node << 1 | 1];
                    seg[node] = (height[l] < height[r]) ? l : r;
                }
            }

            private int query(int node, int b, int e, int L, int R) {
                if (b > R || e < L) {
                    return -1;
                }
                if (b >= L && e <= R) {
                    return seg[node];
                }
                int mid = (b + e) >> 1;
                int left = query(node << 1, b, mid, L, R);
                int right = query(node << 1 | 1, mid + 1, e, L, R);
                if (left == -1) {
                    return right;
                }
                if (right == -1) {
                    return left;
                }
                return height[left] < height[right] ? left : right;
            }
        }
    }

    public static int findLis(int[] a) {
        int lis = 0;
        int[] dp = new int[a.length];
        Arrays.fill(dp, 1);
        for (int i = 0; i < a.length; i++) {
            for (int j = 0; j < i; j++) {
                if (a[j] < a[i]) {
                    dp[i] = Math.max(dp[i], dp[j] + 1);
                }
            }
            lis = Math.max(lis, dp[i]);
        }
        return lis;
    }

    static class SegmentTreeXOR {

        int size;
        int seg[];
        int lazy[];
        int arr[];

        public SegmentTreeXOR(int size, int a[]) {
            this.size = size;
            arr = a;
            seg = new int[size * 4];
            lazy = new int[seg.length];
            build(1, 1, size, a);
        }

        private void build(int idx, int s, int e, int a[]) {
            if (s == e) {
                seg[idx] = a[s];
                return;
            }
            build(idx * 2, s, (s + e) / 2, a);
            build(idx * 2 + 1, (s + e) / 2 + 1, e, a);
            seg[idx] = seg[idx * 2] ^ seg[idx * 2 + 1];
        }

        void process(int idx, int s, int e) {
            if (lazy[idx] == 1) {
                seg[idx] ^= 1;
            }
            if (s < e) {
                lazy[idx * 2] += lazy[idx];
                lazy[idx * 2 + 1] += lazy[idx];
                lazy[idx * 2] %= 2;
                lazy[idx * 2 + 1] %= 2;
            }
            lazy[idx] = 0;
        }

        void updateRange(int idx, int s, int e, int l, int r, int val) {
            process(idx, s, e);
            if ((l > e) || s > r) {
                return;
            }
            if (s >= l && e <= r) {
                lazy[idx] += val;
                lazy[idx] %= 2;
                process(idx, s, e);
                return;
            }
            updateRange(idx * 2, s, (s + e) / 2, l, r, val);
            updateRange(idx * 2 + 1, (s + e) / 2 + 1, e, l, r, val);
            seg[idx] = seg[idx * 2] ^ seg[idx * 2 + 1];
        }

        void updateIndex(int idx, int s, int e, int ind, int val) {
            process(idx, s, e);
            if (ind < s || ind > e) {
                return;
            }
            if (s == ind && ind == e) {
                seg[idx] = val;
                return;
            }
            updateIndex(idx * 2, s, (s + e) / 2, ind, val);
            updateIndex(idx * 2 + 1, (s + e) / 2 + 1, e, ind, val);
            seg[idx] = seg[idx * 2] ^ seg[idx * 2 + 1];
        }

        long sumInRange(int idx, int s, int e, int l, int r) {
            process(idx, s, e);
            if ((l > e) || s > r) {
                return 0;
            }
            if (s >= l && e <= r) {
                return seg[idx];
            }
            return sumInRange(idx * 2, s, (s + e) / 2, l, r) ^ sumInRange(idx * 2 + 1, (s + e) / 2 + 1, e, l, r);
        }
    }

    public static long ceil(long a, long b) {
        return (a + b - 1) / b;
    }

    public static long round(long a, long b) {
        if (a < 0) {
            return (a - b / 2) / b;
        }
        return (a + b / 2) / b;
    }

    static void getPrimeFactors(int a) {
        while (a % 2 == 0) {
            primeFactors.add(2);
            a /= 2;
        }
        for (int i = 3; i * i <= a; i += 2) {
            while (a % i == 0) {
                primeFactors.add(i);
                a /= i;
            }
        }
        if (a > 1) {
            primeFactors.add(a);
        }
    }

    static class SegmentTreeSum {

        int size;
        int seg[];
        int lazy[];
        int arr[];

        public SegmentTreeSum(int size, int a[]) {
            this.size = size;
            arr = a;
            seg = new int[size * 4];
            lazy = new int[size * 4];
            Arrays.fill(lazy, -1);
            build(1, 1, size, a);
        }

        private void build(int idx, int s, int e, int a[]) {
            if (s == e) {
                seg[idx] = a[s];
                return;
            }
            build(idx * 2, s, (s + e) / 2, a);
            build(idx * 2 + 1, (s + e) / 2 + 1, e, a);
            seg[idx] = Math.max(seg[idx * 2], seg[idx * 2 + 1]);

        }

        void process(int idx, int s, int e) {
            if (lazy[idx] == -1) return;
            seg[idx] = lazy[idx];
            if (s < e) {
                lazy[idx * 2] = lazy[idx];
                lazy[idx * 2 + 1] = lazy[idx];
            }
            lazy[idx] = -1;
        }

        void updateRange(int idx, int s, int e, int l, int r, int val) {
            process(idx, s, e);
            if ((l > e) || s > r) {
                return;
            }
            if (s >= l && e <= r) {
                lazy[idx] = val;
                process(idx, s, e);
                return;
            }
            updateRange(idx * 2, s, (s + e) / 2, l, r, val);
            updateRange(idx * 2 + 1, (s + e) / 2 + 1, e, l, r, val);
            seg[idx] = Math.max(seg[idx * 2], seg[idx * 2 + 1]);
        }

        void updateIndex(int idx, int s, int e, int ind, int val) {
            process(idx, s, e);
            if (ind < s || ind > e) {
                return;
            }
            if (s == ind && ind == e) {
                seg[idx] = val;
                return;
            }
            updateIndex(idx * 2, s, (s + e) / 2, ind, val);
            updateIndex(idx * 2 + 1, (s + e) / 2 + 1, e, ind, val);
            seg[idx] = Math.max(seg[idx * 2], seg[idx * 2 + 1]);
        }

        int sumInRange(int idx, int s, int e, int l, int r) {
            process(idx, s, e);
            if ((l > e) || s > r) {
                return Integer.MIN_VALUE;
            }
            if (s >= l && e <= r) {
                return seg[idx];
            }
            return Math.max(sumInRange(idx * 2, s, (s + e) / 2, l, r), sumInRange(idx * 2 + 1, (s + e) / 2 + 1, e, l, r));
        }
    }

    static long mul(long a, long b) {
        return (long) ((long) ((a % mod) * (b % mod)) % mod);
    }

    static class pa {
        int val;
        String name;

        public pa(int val, String name) {
            this.val = val;
            this.name = name;
        }
    }

    static class me {
        String name;
        int cap;
        int pos;

        public me(String name, int cap, int pos) {
            this.name = name;
            this.pos = pos;
            this.cap = cap;
        }
    }

    static boolean can(int servers, int a[], int lim) {
        PriorityQueue<Integer> q = new PriorityQueue<>();
        for (int i = 0; i < servers; i++) {
            q.add(0);
        }
        int idx = 0;
        while (idx < a.length) {
            int time = q.poll();
            if (time + a[idx] > lim) return false;
            q.add(time + a[idx++]);
        }
        return true;
    }

    static int getLog(long v) {
        int cnt = 0;
        while (v > 1) {
            cnt++;
            v >>= 1;
        }
        return cnt;
    }

    static class Node {
        Node node[];
        int val;
        int c;

        public Node() {
            node = new Node[2];
            val = 0;
            c = 0;
        }
    }

    static class XorTrie {

        Node node;

        public XorTrie() {
            node = new Node();
        }

        void insert(int x) {
            Node curr = node;
            for (int i = 30; i >= 0; i--) {
                int val = (x >> i) & 1;
                if (curr.node[val] == null) curr.node[val] = new Node();

                curr = curr.node[val];
                curr.val++;
            }
        }
    }

    static class temp {
        int x;
        String ch;

        public temp(int x, String ch) {
            this.x = x;
            this.ch = ch;
        }
    }

    static int bs(ArrayList<Integer> a, int v) {
        int max = a.size() - 1;
        int min = 0;
        int idx = a.size();
        while (max >= min) {
            int mid = (max + min) / 2;
            if (a.get(mid) >= v) {
                max = mid - 1;
                idx = mid;
            } else {
                min = mid + 1;
            }
        }
        return idx;
    }

    static long getAns(long s, long u, long a[]) {
        long ans = u * (u + 1) / 2;
        if (s > u) {
            long dif = (u - s);
            dif *= -1;
            dif--;
            return ans - (dif) * (dif + 1) / 2;
        } else {
            long dif = u - s;
            return ans - (dif) * (dif + 1) / 2;
        }
    }

    public static long power(long a, long p) {
        long res = 1;
        while (p > 0) {
            if (p % 2 == 0) {
                a = (a * a);
                p /= 2;
            } else {
                res = (res * a);
                p--;
            }
        }
        return res;
    }

    static class Trie {
        Trie fre[];
        int cnt;

        public Trie() {
            fre = new Trie[26];
            cnt = 0;
        }

        public void insert(char ch[], int idx) {
            cnt++;
            if (idx == ch.length) return;
            if (fre[ch[idx] - 'a'] == null) {
                fre[ch[idx] - 'a'] = new Trie();
            }
            fre[ch[idx] - 'a'].insert(ch, idx + 1);
        }
    }

    static pai absSub(pai a, pai b) {
        doWork(a, b);
        pai c = new pai(Math.abs(a.x - b.x), a.y);
        simplifyFraction(b);
        return c;
    }

    static pai add(pai a, pai b) {
        doWork(a, b);
        pai c = new pai(a.x + b.x, a.y);
        simplifyFraction(b);
        return c;
    }

    static pai mu(pai a, pai b) {
        long v = b.x;
        b.x = (int) b.y;
        b.y = v;
        simplifyFraction(b);
        simplifyFraction(a);
        pai c = new pai(a.x * b.x, a.y * b.y);
        simplifyFraction(c);
        return c;
    }

    static long compar(pai a, pai b) {
        doWork(a, b);
        if (a.x < b.x) {
            return -1;
        } else {
            return Math.min(1, a.x - b.x);
        }
    }

    static void doWork(pai a, pai b) {
        simplifyFraction(a);
        simplifyFraction(b);
        long c = a.y;
        a.y *= b.y;
        a.x *= b.y;
        b.y *= c;
        b.x *= c;
    }

    static void simplifyFraction(pai b) {
        long g = GCD(b.x, b.y);
        b.x /= g;
        b.y /= g;
    }

    static int cmp(pair o1, pair o2) {
        if (o1.x > o2.x) {
            return 1;
        } else if (o1.x < o2.x) {
            return -1;
        } else if (o1.y > o2.y) {
            return 1;
        } else if (o1.y < o2.y) {
            return -1;
        }
        return 0;
    }

    public static long LCM(long x, long y) {
        return x / GCD(x, y) * y;
    }

    static class pai {
        int x;
        long y;

        public pai(int x, long y) {
            this.x = x;
            this.y = y;
        }
    }

    static int sum(int x) {
        int s = 0;
        while (x > 0) {
            s += x % 10;
            x /= 10;
        }
        return s;
    }

    static class tm {
        long x;
        int idx;

        public tm(long x, int idx) {
            this.x = x;
            this.idx = idx;
        }
    }

    /*
        static class SegTreeRegularBracket {
            pair seg[];

            public SegTreeRegularBracket(char ch[]) {
                seg = new pair[4 * ch.length];
                build(1, 1, ch.length - 1, ch);
            }

            void build(int idx, int s, int e, char ch[]) {
                if (s == e) {
                    if (ch[s] == '(') {
                        seg[idx] = new pair(1, 0);
                    } else {
                        seg[idx] = new pair(0, 1);
                    }
                    return;
                }
                build(idx << 1, s, (s + e) / 2, ch);
                build(idx << 1 | 1, (s + e) / 2 + 1, e, ch);
                seg[idx] = new pair(seg[idx << 1 | 1].x, seg[idx << 1].y);
                int dif = seg[idx << 1].x - seg[idx << 1 | 1].y;
                int min = Math.min(seg[idx << 1].x, seg[idx << 1 | 1].y);
                seg[idx].d += min * 2 + seg[idx << 1 | 1].d + seg[idx << 1].d;
                if (dif > 0) {
                    seg[idx].x += dif;
                } else {
                    seg[idx].y -= dif;
                }
            }

            pair query(int idx, int s, int e, int l, int r) {
                if (l > e || s > r) {
                    return new pair(0, 0);
                }
                if (s >= l && e <= r) {
                    return seg[idx];
                }
                pair p1 = query(idx << 1, s, (s + e) / 2, l, r);
                pair p2 = query(idx << 1 | 1, (s + e) / 2 + 1, e, l, r);
                pair ans = new pair(p2.x, p1.y);
                int dif = p1.x - p2.y;
                ans.d += p1.d + p2.d;
                ans.d += Math.min(p1.x, p2.y) * 2;
                if (dif > 0) {
                    ans.x += dif;
                } else {
                    ans.y -= dif;
                }
                return ans;
            }
        }

     */
    static class tep {
        int x;
        long y;

        public tep(int x, long y) {
            this.x = x;
            this.y = y;
        }
    }

    static class Tree {
        private ArrayList<Integer> distinctElements;
        private SegmentTree seg;
        private int size;

        public Tree(int a[]) {
            distinctElements = new ArrayList<>();
            Arrays.sort(a);
            distinctElements.add(a[0]);
            for (int i = 1; i < a.length; i++) {
                if (a[i] != distinctElements.get(distinctElements.size() - 1)) {
                    distinctElements.add(a[i]);
                }
            }
            size = distinctElements.size() + 5;
            seg = new SegmentTree(size);
        }

        public void add(int val) {
            int idx = getIndex(val);
            seg.updateIndex(1, 1, size, idx, 1);
        }

        public void remove(int val) {
            int idx = getIndex(val);
            seg.updateIndex(1, 1, size, idx, -1);
        }

        public int getKth(int val) {
            return distinctElements.get(seg.getKth(1, 1, size, val) - 1);
        }

        private int getIndex(int val) {
            int max = distinctElements.size() - 1;
            int min = 0;
            int ans = -1;
            while (max >= min) {
                int mid = (max + min) / 2;
                if (distinctElements.get(mid) == val) {
                    return mid + 2;
                } else if (distinctElements.get(mid) > val) {
                    max = mid - 1;
                } else {
                    min = mid + 1;
                }
            }
            return ans;
        }

        public int countNumbersGreaterThan(int val) {
            int idx = getIndex(val);
            return seg.sumInRange(1, 1, size, idx + 1, size);
        }

        public int countNumbersLessThan(int val) {
            int idx = getIndex(val);
            return seg.sumInRange(1, 1, size, 1, idx - 1);
        }

        private static class SegmentTree {

            int size;
            int seg[];
            int arr[];

            public SegmentTree(int size) {
                this.size = size;
                seg = new int[size * 4];
            }

            void updateIndex(int idx, int s, int e, int ind, int val) {
                if (ind < s || ind > e) {
                    return;
                }
                if (s == ind && ind == e) {
                    seg[idx] += val;
                    return;
                }
                updateIndex(idx << 1, s, (s + e) / 2, ind, val);
                updateIndex(idx << 1 | 1, (s + e) / 2 + 1, e, ind, val);
                seg[idx] = seg[idx << 1] + seg[idx << 1 | 1];
            }

            int getKth(int idx, int s, int e, int val) {
                if (s == e) {
                    return e;
                }
                if (seg[idx << 1] >= val) {
                    return getKth(idx << 1, s, (s + e) / 2, val);
                }
                return getKth(idx << 1 | 1, (s + e) / 2 + 1, e, val - seg[idx << 1]);
            }

            int sumInRange(int idx, int s, int e, int l, int r) {
                if ((l > e) || s > r) {
                    return 0;
                }
                if (s >= l && e <= r) {
                    return seg[idx];
                }
                return sumInRange(idx << 1, s, (s + e) / 2, l, r) + sumInRange(idx << 1 | 1, (s + e) / 2 + 1, e, l, r);
            }
        }
    }

    static void prefixSum2D(int arr[][]) {
        for (int i = 0; i < arr.length; i++) {
            prefixSum(arr[i]);
        }
        for (int i = 0; i < arr[0].length; i++) {
            for (int j = 1; j < arr.length; j++) {
                arr[j][i] += arr[j - 1][i];
            }
        }
    }

    static int sumOfRange(int x1, int y1, int x2, int y2, int a[][]) {
        return (a[x2][y2] - a[x1 - 1][y2] - a[x2][y1 - 1]) + a[x1 - 1][y1 - 1];
    }

    static class SparseTableMin {
        int table[][], logTable[], powers[];
        int a[];

        public SparseTableMin(int a[]) {
            this.a = a;
            logTable = new int[a.length + 1];
            for (int i = 2; i <= a.length; i++) {
                logTable[i] = logTable[i / 2] + 1;
            }
            int maxLog = logTable[a.length] + 1;
            table = new int[a.length][maxLog];
            powers = new int[maxLog];
            powers[0] = 1;
            for (int i = 1; i < powers.length; i++) {
                powers[i] = powers[i - 1] * 2;
            }
            for (int i = 0; i < a.length; i++) {
                table[i][0] = a[i];
            }
            for (int i = 1; i < maxLog; i++) {
                for (int j = 0; j + powers[i] - 1 < a.length; j++) {
                    table[j][i] = Math.min(table[j][i - 1], table[j + powers[i - 1]][i - 1]);
                }
            }
        }

        int query(int l, int r) {
            int log = logTable[r - l + 1];
            return Math.min(table[l][log], table[r - powers[log] + 1][log]);
        }
    }

    static class SparseTableGCD {
        long table[][];
        int logTable[], powers[];
        int a[];

        public SparseTableGCD(int a[]) {
            this.a = a;
            logTable = new int[a.length + 1];
            for (int i = 2; i <= a.length; i++) {
                logTable[i] = logTable[i / 2] + 1;
            }
            int maxLog = logTable[a.length] + 1;
            table = new long[a.length][maxLog];
            powers = new int[maxLog];
            powers[0] = 1;
            for (int i = 1; i < powers.length; i++) {
                powers[i] = powers[i - 1] * 2;
            }
            for (int i = 0; i < a.length; i++) {
                table[i][0] = a[i];
            }
            for (int i = 1; i < maxLog; i++) {
                for (int j = 0; j + powers[i] - 1 < a.length; j++) {
                    table[j][i] = GCD(table[j][i - 1], table[j + powers[i - 1]][i - 1]);
                }
            }
        }

        long query(int l, int r) {
            int log = logTable[r - l + 1];
            return GCD(table[l][log], table[r - powers[log] + 1][log]);
        }
    }

    static class SparseTable {
        int table[][], logTable[], powers[];
        int a[];

        public SparseTable(int a[]) {
            this.a = a;
            logTable = new int[a.length + 1];
            for (int i = 2; i <= a.length; i++) {
                logTable[i] = logTable[i / 2] + 1;
            }
            int maxLog = logTable[a.length] + 1;
            table = new int[a.length][maxLog];
            powers = new int[maxLog];
            powers[0] = 1;
            for (int i = 1; i < powers.length; i++) {
                powers[i] = powers[i - 1] * 2;
            }
            for (int i = 0; i < a.length; i++) {
                table[i][0] = a[i];
            }
            for (int i = 1; i < maxLog; i++) {
                for (int j = 0; j + powers[i] - 1 < a.length; j++) {
                    table[j][i] = Math.max(table[j][i - 1], table[j + powers[i - 1]][i - 1]);
                }
            }
        }

        int query(int l, int r) {
            int log = logTable[r - l + 1];
            return Math.max(table[l][log], table[r - powers[log] + 1][log]);
        }
    }

    static class tem {
        int val;
        long sum;
        int idx;

        public tem(int val, long sum, int idx) {
            this.val = val;
            this.sum = sum;
            this.idx = idx;
        }
    }

    static void tarjan(int node, int par) {
        num[node] = lowesLink[node] = nu++;
        st.add(node);
        inStack[node] = true;
        for (Integer ch : g[node]) {
            if (num[ch] == 0) {
                tarjan(ch, node);
                lowesLink[node] = Math.min(lowesLink[node], lowesLink[ch]);
            } else if (inStack[ch]) {
                lowesLink[node] = Math.min(lowesLink[node], num[ch]);
            }
        }
        if (lowesLink[node] == num[node]) {
            int x = -1;
            stronConCom.add(new ArrayList<>());
            while (x != node) {
                x = st.pop();
                inStack[x] = false;
                stronConCom.getLast().add(x);
            }
        }
    }

    static long add(long a, long b) {
        a += b;
        if (a >= mod) {
            a -= mod;
        }
        return a;
    }

    static long modinv(long x) {
        return fast_pow(x, mod - 2, mod);
    }

    static long Div(long x, long y) {
        return mul(x, modinv(y));

    }

    static long mod(long a, long b) {
        long r = a % b;
        return r < 0 ? r + b : r;
    }

    static long Sub(long x, long y) {
        long z = x - y;
        if (z < 0) {
            z += mod;
        }
        return z;
    }

    static long fast_pow(long a, long p, long mod) {
        long res = 1;
        while (p > 0) {
            if (p % 2 == 0) {
                a = ((a % mod) * (a % mod)) % mod;
                p /= 2;
            } else {
                res = ((res % mod) * (a % mod)) % mod;
                p--;
            }
        }
        return res;
    }

    public static pair[] dijkstra(int node, ArrayList<pair> a[]) {
        PriorityQueue<tri> q = new PriorityQueue<>(new Comparator<tri>() {
            @Override
            public int compare(tri o1, tri o2) {
                if (o1.y > o2.y) {
                    return 1;
                } else if (o1.y < o2.y) {
                    return -1;
                } else {
                    return 0;
                }
            }
        });
        q.add(new tri(node, 0, -1));
        pair distance[] = new pair[a.length];
        while (!q.isEmpty()) {
            tri p = q.poll();
            int cost = p.y;
            if (distance[p.x] != null) {
                continue;
            }
            distance[p.x] = new pair(p.pa, cost);
            ArrayList<pair> nodes = a[p.x];
            for (pair node1 : nodes) {
                if (distance[node1.x] == null) {
                    tri pa = new tri(node1.x, cost + node1.y, p.x);
                    q.add(pa);
                }
            }
        }
        return distance;
    }

    static class tri {

        int x;
        int y;
        int pa;

        public tri(int x, int y, int pa) {
            this.x = x;
            this.y = y;
            this.pa = pa;
        }

    }

    static void dfs(int node, ArrayList<Integer> g[], ArrayList<Integer> con, boolean vi[]) {
        vi[node] = true;
        con.add(node);
        for (Integer x : g[node]) {
            if (!vi[x]) {
                dfs(x, g, con, vi);
            }
        }
    }

    static long sqrt(long v) {
        long max = (long) 4e9;
        long min = 0;
        long ans = 0;
        while (max >= min) {
            long mid = (max + min) / 2;
            if (mid * mid > v) {
                max = mid - 1;
            } else {
                ans = mid;
                min = mid + 1;
            }
        }
        return ans;
    }

    static long cbrt(long v) {
        long max = (long) 3e6;
        long min = 0;
        long ans = 0;
        while (max >= min) {
            long mid = (max + min) / 2;
            if (mid * mid > v) {
                max = mid - 1;
            } else {
                ans = mid;
                min = mid + 1;
            }
        }
        return ans;
    }

    static int revI(int n) {
        int re = 0;
        while (n > 0) {
            re = re * 10 + n % 10;
            n /= 10;
        }
        return re;
    }

    static class Edge {
        int u, v;
        long cost;

        public Edge(int u, int v, long cost) {
            this.u = u;
            this.v = v;
            this.cost = cost;
        }

        public String toString() {
            return u + " " + v + " " + cost;
        }
    }

    static class SegmentTree {

        int size;
        int seg[];
        int arr[];

        public SegmentTree(int size, int a[]) {
            this.size = size;
            seg = new int[size * 4];
            Arrays.fill(seg, 1);
            build(1, 1, size, a);
        }

        private void build(int idx, int s, int e, int a[]) {
            if (s == e) {
                seg[idx] = a[s];
                return;
            }
            build(idx << 1, s, (s + e) / 2, a);
            build(idx << 1 | 1, (s + e) / 2 + 1, e, a);
            seg[idx] = seg[idx << 1] + seg[idx << 1 | 1];
        }

        void updateIndex(int idx, int s, int e, int ind, int val) {
            if (ind < s || ind > e) {
                return;
            }
            if (s == ind && ind == e) {
                seg[idx] += val;
                return;
            }
            updateIndex(idx << 1, s, (s + e) / 2, ind, val);
            updateIndex(idx << 1 | 1, (s + e) / 2 + 1, e, ind, val);
            seg[idx] = seg[idx << 1] + seg[idx << 1 | 1];
        }

        int sumInRange(int idx, int s, int e, int l, int r) {
            if ((l > e) || s > r) {
                return 1;
            }
            if (s >= l && e <= r) {
                return seg[idx];
            }
            return (sumInRange(idx << 1, s, (s + e) / 2, l, r) + sumInRange(idx << 1 | 1, (s + e) / 2 + 1, e, l, r));
        }
    }

    public static void prefixSum(long[] a) {
        for (int i = 1; i < a.length; i++) {
            a[i] += a[i - 1];
        }
    }

    public static void suffixSum(int[] a) {
        for (int i = a.length - 2; i > -1; i--) {
            a[i] += a[i + 1];
        }
    }

    public static void prefixSum(int[] a) {
        for (int i = 1; i < a.length; i++) {
            a[i] = a[i] + a[i - 1];
        }
    }

    public static void suffixSum(long[] a) {
        for (int i = a.length - 2; i > -1; i--) {
            a[i] = a[i] + a[i + 1];
        }
    }

    static class HashedString {
        // Change M and B if you want
        public static final long M = 911382323;
        public static final long B = 972663749;

        // pow[i] contains B^i % M
        private static ArrayList<Long> pow = new ArrayList<>();

        // pHash[i] is the hash of the first i characters of the given string
        private long[] pHash;

        public HashedString(String s) {
            if (pow.isEmpty()) {
                pow.add(1L);
            }
            while (pow.size() <= s.length()) {
                pow.add((pow.get(pow.size() - 1) * B) % M);
            }
            pHash = new long[s.length() + 1];
            pHash[0] = 0;
            for (int i = 0; i < s.length(); i++) {
                pHash[i + 1] = ((pHash[i] * B) % M + s.charAt(i)) % M;
            }
        }

        public long getHash(int start, int end) {
            long rawVal = pHash[end + 1] - (pHash[start] * pow.get(end - start + 1));
            return (rawVal % M + M) % M;
        }
    }

    static class SegmentTreeMax {

        int size;
        int seg[];
        int lazy[];
        long arr[];

        public SegmentTreeMax(int size, long[] a) {
            this.size = size;
            arr = a;
            seg = new int[size * 4];
            //lazy = new int[size * 4];
            build(1, 1, size, a);
        }

        private void build(int idx, int s, int e, long a[]) {
            if (s == e) {
                seg[idx] = s;
                return;
            }
            build(idx * 2, s, (s + e) / 2, a);
            build(idx * 2 + 1, (s + e) / 2 + 1, e, a);
            if (arr[seg[idx * 2 + 1]] <= arr[seg[idx * 2]]) {
                seg[idx] = seg[idx * 2];
            } else {
                seg[idx] = seg[idx * 2 + 1];
            }
        }

        void process(int idx, int s, int e) {
            seg[idx] += lazy[idx];
            if (s < e) {
                lazy[idx * 2] += lazy[idx];
                lazy[idx * 2 + 1] += lazy[idx];
            }
            lazy[idx] = 0;
        }

        void updateRange(int idx, int s, int e, int l, int r, int val) {
            process(idx, s, e);
            if ((l > e) || s > r) {
                return;
            }
            if (s >= l && e <= r) {
                lazy[idx] += val;
                process(idx, s, e);
                return;
            }
            updateRange(idx * 2, s, (s + e) / 2, l, r, val);
            updateRange(idx * 2 + 1, (s + e) / 2 + 1, e, l, r, val);
            if (arr[seg[idx * 2 + 1]] <= arr[seg[idx * 2]]) {
                seg[idx] = seg[idx * 2];
            } else {
                seg[idx] = seg[idx * 2 + 1];
            }
        }

        void updateIndex(int idx, int s, int e, int ind, int val) {
            if (ind < s || ind > e) {
                return;
            }
            if (s == ind && ind == e) {
                arr[s] = val;
                return;
            }
            updateIndex(idx * 2, s, (s + e) / 2, ind, val);
            updateIndex(idx * 2 + 1, (s + e) / 2 + 1, e, ind, val);
            if (arr[seg[idx * 2 + 1]] <= arr[seg[idx * 2]]) {
                seg[idx] = seg[idx * 2];
            } else {
                seg[idx] = seg[idx * 2 + 1];
            }
        }

        int maxInRange(int idx, int s, int e, int l, int r) {
            //  process(idx, s, e);
            if ((l > e) || s > r) {
                return Integer.MIN_VALUE;
            }
            if (s >= l && e <= r) {
                return seg[idx];
            }
            int a1 = maxInRange(idx * 2, s, (s + e) / 2, l, r);
            int a2 = maxInRange(idx * 2 + 1, (s + e) / 2 + 1, e, l, r);
            if (a1 == Integer.MIN_VALUE) return a2;
            if (a2 == Integer.MIN_VALUE) return a1;
            if (arr[a1] >= arr[a2]) return a1;
            return a2;
        }
    }

    static class SegmentTreeMin {
        static class Node {
            int val, cnt;

            public Node(int val, int cnt) {
                this.val = val;
                this.cnt = cnt;
            }
        }

        int size;
        int seg[];
        int lazy[];
        long arr[];

        public SegmentTreeMin(int size, long a[]) {
            this.size = size;
            arr = a;
            seg = new int[size * 4];
            //  lazy = new int[size * 4];
            build(1, 1, size, a);
        }

        private void build(int idx, int s, int e, long a[]) {
            if (s == e) {
                seg[idx] = s;
                return;
            }
            build(idx * 2, s, (s + e) / 2, a);
            build(idx * 2 + 1, (s + e) / 2 + 1, e, a);
            if (arr[seg[idx * 2 + 1]] >= arr[seg[idx * 2]]) {
                seg[idx] = seg[idx * 2];
            } else {
                seg[idx] = seg[idx * 2 + 1];
            }
        }

        void process(int idx, int s, int e) {
            /*
            seg[idx] += lazy[idx];
            if (s < e) {
                lazy[idx * 2] += lazy[idx];
                lazy[idx * 2 + 1] += lazy[idx];
            }
            lazy[idx] = 0;

             */
        }

        void updateIndex(int idx, int s, int e, int ind, int val) {
            if (ind < s || ind > e) {
                return;
            }
            if (s == ind && ind == e) {
                arr[s] = val;
                return;
            }
            updateIndex(idx * 2, s, (s + e) / 2, ind, val);
            updateIndex(idx * 2 + 1, (s + e) / 2 + 1, e, ind, val);
            if (arr[seg[idx * 2 + 1]] >= arr[seg[idx * 2]]) {
                seg[idx] = seg[idx * 2];
            } else {
                seg[idx] = seg[idx * 2 + 1];
            }
        }

        void updateRange(int idx, int s, int e, int l, int r, int val) {
            process(idx, s, e);
            if ((l > e) || s > r) {
                return;
            }
            if (s >= l && e <= r) {
                lazy[idx] += val;
                process(idx, s, e);
                return;
            }
            updateRange(idx * 2, s, (s + e) / 2, l, r, val);
            updateRange(idx * 2 + 1, (s + e) / 2 + 1, e, l, r, val);
            if (arr[seg[idx * 2 + 1]] >= arr[seg[idx * 2]]) {
                seg[idx] = seg[idx * 2];
            } else {
                seg[idx] = seg[idx * 2 + 1];
            }
        }

        int minInRange(int idx, int s, int e, int l, int r) {
            //  process(idx, s, e);
            if (l > e || s > r) {
                return Integer.MAX_VALUE;
            }
            if (s >= l && e <= r) {
                return seg[idx];
            }
            int a1 = minInRange(idx * 2, s, (s + e) / 2, l, r);
            int a2 = minInRange(idx * 2 + 1, (s + e) / 2 + 1, e, l, r);
            if (a1 == Integer.MAX_VALUE) return a2;
            if (a2 == Integer.MAX_VALUE) return a1;
            if (arr[a1] <= arr[a2]) return a1;
            return a2;
        }
    }

    static boolean isPalindrome(String w) {
        for (int i = 0; i < w.length() / 2; i++) {
            if (w.charAt(i) != w.charAt(w.length() - i - 1)) return false;
        }
        return true;
    }

    public static boolean isValid(int i, int j, int n, int m) {
        return (i > -1 && i < n) && (j > -1 && j < m);

    }

    static int cmpDouble(double a, double b) {
        if (Math.abs(a - b) <= 1e-7) {
            return 0;
        }
        return a < b ? -1 : 1;
    }

    public static long GCD(long x, long y) {
        while (y != 0) {
            long c = x % y;
            x = y;
            y = c;
        }
        return x;
    }

    static boolean isPrime(long num) {
        if (num <= 1) {
            return false;
        }
        if (num == 2) {
            return true;
        }
        if (num % 2 == 0) {
            return false;
        }
        if (num == 3) {
            return true;
        }
        for (long i = 3; i * i <= num; i += 2) {
            if (num % i == 0) {
                return false;
            }
        }
        return true;
    }

    public static int[] swap(int data[], int left, int right) {

        // Swap the data
        int temp = data[left];
        data[left] = data[right];
        data[right] = temp;

        // Return the updated array
        return data;
    }

    public static int[] reverse(int data[], int left, int right) {

        // Reverse the sub-array
        while (left < right) {
            int temp = data[left];
            data[left++] = data[right];
            data[right--] = temp;
        }

        // Return the updated array
        return data;
    }

    public static boolean findNextPermutation(int data[]) {

        if (data.length <= 1) {
            return false;
        }

        int last = data.length - 2;

        // find the longest non-increasing suffix
        // and find the pivot
        while (last >= 0) {
            if (data[last] < data[last + 1]) {
                break;
            }
            last--;
        }

        // If there is no increasing pair
        // there is no higher order permutation
        if (last < 0) {
            return false;
        }

        int nextGreater = data.length - 1;

        // Find the rightmost successor to the pivot
        for (int i = data.length - 1; i > last; i--) {
            if (data[i] > data[last]) {
                nextGreater = i;
                break;
            }
        }

        // Swap the successor and the pivot
        data = swap(data, nextGreater, last);

        // Reverse the suffix
        data = reverse(data, last + 1, data.length - 1);

        // Return true as the next_permutation is done
        return true;
    }

    static double calcArea(pair a, pair b, pair c) {
        double ret = Math.abs((a.x * (double) (b.y - c.y)) + (b.x * (double) (c.y - a.y)) + (c.x * (double) (a.y - b.y))) / 2.0;
        return ret;
    }

    static class MultiSet<T> {

        TreeMap<T, Integer> fre;
        TreeSet<T> set;
        int size;

        public MultiSet() {
            set = new TreeSet<>();
            fre = new TreeMap<>();
            size = 0;
        }

        public MultiSet(Comparator<T> cmp) {
            set = new TreeSet<>(cmp);
            fre = new TreeMap<>(cmp);
            size = 0;
        }

        public void add(T elem) {
            if (fre.get(elem) == null || fre.get(elem) == 0) {
                fre.put(elem, 0);
                set.add(elem);
            }
            fre.put(elem, fre.get(elem) + 1);
            size++;
        }

        public void remove(T elem) {
            if (!set.contains(elem)) return;
            fre.put(elem, fre.get(elem) - 1);
            if (fre.get(elem) == 0) {
                set.remove(elem);
            }
            size--;
        }

        public boolean contains(T elem) {
            return set.contains(elem);
        }
    }

    static class Reader extends PrintWriter {

        private BufferedReader r;
        private StringTokenizer st;
        // standard input

        public Reader() {
            this(System.in, System.out);
        }

        public Reader(InputStream i, OutputStream o) {
            super(o);
            r = new BufferedReader(new InputStreamReader(i));
        }
        // USACO-style file input

        public Reader(String problemName) throws IOException {
            super(problemName + ".out");
            r = new BufferedReader(new FileReader(problemName));
        }

        // returns null if no more input
        String nextLine() {
            String str = "";
            try {
                str = r.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }

        public String next() {
            try {
                while (st == null || !st.hasMoreTokens()) {
                    st = new StringTokenizer(r.readLine());
                }
                return st.nextToken();
            } catch (Exception e) {
            }
            return null;
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }

        public long nextLong() {

            return Long.parseLong(next());
        }
    }

    static class pair {
        public int x, d, y;
        public long c;

        public pair(int x, int y) {
            this.x = x;
            this.y = y;
            this.d = d;
        }

        public String toString() {
            return x + " " + y;
        }
    }

    static class DSU {
        int[] parent, groupSize;
        int numberOfNodes, numberOfGroups, maxGroup;

        public DSU(int numberOfNodes) {
            this.numberOfNodes = numberOfNodes;
            parent = new int[numberOfNodes + 1];
            groupSize = new int[numberOfNodes + 1];
            numberOfGroups = numberOfNodes;
            maxGroup = 1;
            for (int i = 1; i <= numberOfNodes; i++) {
                parent[i] = i;
                groupSize[i] = 1;
            }
        }

        public int getLeader(int x) {
            return parent[x] = (parent[x] == x ? x : getLeader(parent[x]));
        }

        public boolean sameGroup(int x, int y) {
            return getLeader(x) == getLeader(y);
        }

        public long mergeGroups(int x, int y) {
            int leader1 = getLeader(x);
            int leader2 = getLeader(y);
            if (leader1 != leader2) {
                numberOfGroups--;
                if (groupSize[leader1] < groupSize[leader2]) {
                    int temp = leader1;
                    leader1 = leader2;
                    leader2 = temp;
                }
                parent[leader2] = leader1;
                long v = groupSize[leader2] * (long) (groupSize[leader2] - 1) / 2;
                v += groupSize[leader1] * (long) (groupSize[leader1] - 1) / 2;
                groupSize[leader1] += groupSize[leader2];
                maxGroup = Math.max(maxGroup, groupSize[leader1]);
                return groupSize[leader1] * (long) (groupSize[leader1] - 1) / 2 - v;
            } else {
                return 0;
            }
        }

        public int getSize(int x) {
            return groupSize[getLeader(x)];
        }
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios that might cause the algorithm to fail. Here are the steps to create such a test input:

1. **Understand the Problem**: The problem involves an array of integers and finding the index that minimizes the absolute difference between the sum of the first part and the negative sum of the second part of the array.

2. **Identify Edge Cases**:
   - **Large Values for n and k**: The code should handle large values of \( n \) and \( k \) efficiently.
   - **Small Values for n and k**: Test with small values to ensure the algorithm works for simple cases.
   - **Boundary Conditions**: Test with \( n \) and \( k \) at the boundary of their ranges.
   - **Special Cases**: Test with values where the array is symmetric or where the sum is zero.

3. **Generate Test Input**:
   - **Test Case 1**: Large values for \( n \) and \( k \).
   - **Test Case 2**: Small values for \( n \) and \( k \).
   - **Test Case 3**: Boundary values for \( n \) and \( k \).
   - **Test Case 4**: Special case where the array is symmetric or the sum is zero.

Based on these considerations, here is a small-scale test input that covers these cases:

```plaintext
4
2 2
7 2
5 3
1000000000 1000000000
```

This test input includes:
- A small array with \( n = 2 \) and \( k = 2 \).
- A medium-sized array with \( n = 7 \) and \( k = 2 \).
- Another medium-sized array with \( n = 5 \) and \( k = 3 \).
- A large array with \( n = 1000000000 \) and \( k = 1000000000 \).

These cases will help in verifying the correctness and efficiency of the algorithm.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases.

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
#include <bits/stdc++.h>
#define ll long long
#define lli long long int
#define nl "\n"
#define fl(x) for (int i = 0; i < x; i++)
#define fl1(x) for (int i = 1; i < x; i++)
#define sv(v) sort(v.begin(), v.end())
#define vi vector<int>
#define vout(v)      \
    for (auto i : v) \
    cout << i << " "
#define pp pair<int, int>
#define co(n) cout << n

using namespace std;

ll calculateSum(ll start, ll end) {
    return (end * (end + 1)) / 2 - (start * (start - 1)) / 2;
}

ll findMinimumDifference(ll sum, ll c) {
    ll mn = INT64_MAX;
    ll s = 0;
    while (true) {
        s += c;
        sum -= c;
        c--;
        ll ans = abs(sum - s);
        mn = min(mn, abs(sum - s));
        if (ans > mn)
            break;
    }
    return mn;
}

void solve() {
    ll a, b;
    cin >> a >> b;
    ll c = b + a - 1;
    ll sum = calculateSum(b, c);
    ll mn = findMinimumDifference(sum, c);
    cout << mn << nl;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);

    int t;
    cin >> t;
    while (t--)
        solve();

    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
1000000000 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases.

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
#include <bits/stdc++.h>

using namespace std;
#define ll            long long
#define int            long long int
#define F              first
#define S              second
#define pb             push_back
#define si             set <int>
#define vi             vector <int>
#define pii            pair <int, int>
#define vpi            vector <pii>
#define vpp            vector <pair<int, pii> >
#define mii            map <int, int>
#define mpi            map <pii, int>
#define spi            set <pii>
#define endl           "\n"
#define sz(x)          ((int) x.size())
#define all(p)         p.begin(), p.end()
#define double         long double
#define maxpq          priority_queue <int>
#define minpq          priority_queue <int, vi, greater<int> >
#define print(a)       for(auto x : a) cout << x << " "; cout << endl
#define print1(a)      for(auto x : a) cout << x.F << " " << x.S << endl
#define print2(a,x,y)  for(int i = x; i < y; i++) cout<< a[i]<< " "; cout << endl
#define printM(arr)    for(auto &a:arr) {for(int &b:a){cout<<b<<" ";} cout<<endl;}
#define yy "YES\n"
#define nn "NO\n"
#define pp(x)          cout<<x<<endl;
#define fill(arr,n)      for(int i=0;i<n;i++){int d;cin>>d;arr.push_back(d);}
#define forn(n)            for(int i=0;i<n;i++)
const int N = 200005,M=1e9+7;
inline int power(int a, int b)
{
  int x = 1;
  while (b)
  {
    if (b & 1) x *= a;
    a *= a;
    b >>= 1;
  }
  return x;
}

void solve() {
  // int a, b;
  // cin>>a>>b;
  // cout<<b-a<<endl;

  // int n;
  // cin>>n;

  // vector<int>ans;
  // for(int i = 0; i<n; i++){
  //   string s;
  //   cin>>s;
  //   // cout<<s;
  //   for(int j = 0; j<4; j++){
  //     if(s[j] == '#'){
  //       ans.push_back(j+1);
  //       break;
  //     }
  //   }
  // }

  // for(int i = n-1; i>=0; i--)cout<<ans[i]<<" ";
  //   cout<<endl;

  // int x,y,k;
  // cin>>x>>y>>k;

  // int a = (x+k-1)/k;
  // int b = (y+k-1)/k;

  // int ans = 2*(min(a, b) + abs(b-a));

  // if(a>b){
  //   ans --;
  // }
  // cout<<ans<<endl;

  int n,k;
  cin>>n>>k;

  int total = (k+n-1)*(k+n)/2;
  int sub = (k-1)*k/2;
  total -= sub;

  int ans = n+k;
  // int ps = 1e18;
  for(int i = n+k-1; i>=k; i--){
      int fs = i*(i+1)/2 - sub;
      int ns = abs(total- 2*fs);
      // cout<<ns<<endl;
      // if(ns>ps)break;
      // ps = ns;
      ans = min(ans, ns);
  }

  cout<<ans<<endl;
}

int32_t main()
{
  ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#ifndef ONLINE_JUDGE
  freopen("input.txt",  "r",  stdin);
  freopen("output.txt", "w", stdout);
#endif
  int t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that can cause the algorithm to fail. The code provided has a few potential issues:

1. **Overflow**: The code uses `int` (which is `long long int` in this case) for calculations, but the values can still be very large, leading to potential overflow issues.
2. **Incorrect Calculation of `total` and `sub`**: The calculation of `total` and `sub` might not be correct, especially for very large values of `n` and `k`.
3. **Looping and Condition Checks**: The loop and condition checks might not be optimal and can lead to incorrect results for certain values of `n` and `k`.

Let's consider the following edge cases:

1. **Large `n` and `k`**: Test the code with very large values of `n` and `k` to check for overflow.
2. **Small `n` and `k`**: Test with small values to ensure the logic works correctly for simple cases.
3. **Boundary Values**: Test with boundary values of `n` and `k` to ensure the code handles edge cases correctly.

Here is a small-scale test input that includes these edge cases:

```plaintext
4
2 2
7 2
5 3
1000000000 1000000000
```

This test input includes:
- A small case with `n = 2` and `k = 2` to ensure the basic logic works.
- A case with `n = 7` and `k = 2` to test a slightly larger array.
- A case with `n = 5` and `k = 3` to test another small but different case.
- A large case with `n = 1000000000` and `k = 1000000000` to test for potential overflow and performance issues.

This should help in exposing any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases.

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
t=int(input())
for i in range(t):
	n,k=map(int, input().split())
	a=(n-n*n)/2
	j=1+4*(k*k-a-k+k*n)
	x=int(j**0.5-1)//2
	print(int(min(abs((x*(x+1)-k*(k-1)-(k+n-1)*(k+n)+x*(x+1))/2), abs(((x+1)*(x+2)-k*(k-1))/2 -((k+n-1)*(k+n)-(x+1)*(x+2))/2))))

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The buggy code has several potential issues, such as incorrect calculations, integer overflow, and handling large values improperly.

Here are the steps to identify and create such a test input:

1. **Edge Cases:**
   - Small values of \( n \) and \( k \).
   - Large values of \( n \) and \( k \) to test for integer overflow.
   - Values where the sum of elements is very close to zero.

2. **Large Values:**
   - \( n \) and \( k \) both close to the upper limit of \( 10^9 \).

3. **Small Values:**
   - \( n \) and \( k \) both close to the lower limit of 2.

4. **Balanced Values:**
   - Values where the sum of the first half and the second half of the array are close to each other.

Given these considerations, a small-scale test input that can expose the bug is:

```plaintext
3
2 2
1000000000 1000000000
500000000 500000000
```

This input includes:
- A small case where \( n = 2 \) and \( k = 2 \) to test the basic functionality.
- A large case where \( n \) and \( k \) are both \( 10^9 \) to test for integer overflow and large value handling.
- A balanced case where \( n \) and \( k \) are both \( 5 \times 10^8 \) to test for values where the sum of elements is close to zero.

This should help expose any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases.

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
def value(k,i,n):
    v = i*(i-1)//2
    t= n*(n-1)//2

    return(abs(k*(2*i-n) + (2*v-t)))

for _ in range(int(input())):
    n,k = [int(x) for x in input().split()]
    left= 0
    right = n
    minvalue = 1e9

    while(left<=right):
        mid = left+ (right-left)//2

        value1 = value(k,mid,n)
        value2 = value(k,mid+1,n)
        value3 = value(k,mid-1,n)

        if value1<=value2 and value1<=value3:
            minvalue= min(minvalue, value1)
            break
        elif value1>value2:
            minvalue= min(minvalue,value2)
            left= mid+1
        else:
            minvalue = min(minvalue,value3)
            right= mid-1
    print(minvalue)

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. The buggy code uses a binary search approach to find the minimum value of \( x \). The main issue with the code is that it might not handle large values of \( n \) and \( k \) correctly, especially when the differences between \( value1 \), \( value2 \), and \( value3 \) are very small or when the values are very large.

Here are some key points to consider:
1. **Large Values of \( n \) and \( k \)**: The code should handle very large values of \( n \) and \( k \) without overflowing or producing incorrect results.
2. **Edge Cases**: Consider the smallest and largest possible values for \( n \) and \( k \).
3. **Symmetry and Boundary Conditions**: Test cases where the array is symmetric or the index \( i \) is at the boundary can be tricky.

Let's create a small-scale test input that includes a mix of these scenarios:

```plaintext
5
2 1
3 1
4 1
1000000000 1
1000000000 1000000000
```

### Explanation:
1. **Case 1: \( n = 2, k = 1 \)**
   - Array: \([1, 2]\)
   - Expected output: \(1\)

2. **Case 2: \( n = 3, k = 1 \)**
   - Array: \([1, 2, 3]\)
   - Expected output: \(1\)

3. **Case 3: \( n = 4, k = 1 \)**
   - Array: \([1, 2, 3, 4]\)
   - Expected output: \(2\)

4. **Case 4: \( n = 1000000000, k = 1 \)**
   - Array: \([1, 2, \ldots, 1000000000]\)
   - Expected output: \(500000000\)

5. **Case 5: \( n = 1000000000, k = 1000000000 \)**
   - Array: \([1000000000, 1000000001, \ldots, 1999999999]\)
   - Expected output: \(500000000000000000\)

This test input includes both small and large values, and it tests the edge cases and symmetry of the array. It should help expose any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases.

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
import java.util.*;

public class A3 {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- != 0) {
            long n = sc.nextLong();
            long k = sc.nextLong();

            long sum = n*k+ (n)*(n-1)/2;

            long min = sum;
            for (long i = n-1; i >=0; i--) {
                sum -= 2 * (k + i);
                if(min>Math.abs(sum))min=Math.abs(sum);
                else break;
            }
            System.out.println(min);
        }
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
2 1000000000
1000000000 1
1000000000 1000000000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Klee's SUPER DUPER LARGE Array!!!
time_limit_ms: 2000
memory_limit_mb: 256
Description: Klee has an array $$$a$$$ of length $$$n$$$ containing integers $$$[k, k+1, ..., k+n-1]$$$ in that order. Klee wants to choose an index $$$i$$$ ($$$1 \leq i \leq n$$$) such that $$$x = |a_1 + a_2 + \dots + a_i - a_{i+1} - \dots - a_n|$$$ is minimized. Note that for an arbitrary integer $$$z$$$, $$$|z|$$$ represents the absolute value of $$$z$$$.

Output the minimum possible value of $$$x$$$.

Input: The first line contains $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases.

Each test case contains two integers $$$n$$$ and $$$k$$$ ($$$2 \leq n, k \leq 10^9$$$) — the length of the array and the starting element of the array.

Output: For each test case, output the minimum value of $$$x$$$ on a new line.

Examples:
input:
4
2 2
7 2
5 3
1000000000 1000000000
output:
1
5
1
347369930

Note: In the first sample, $$$a = [2, 3]$$$. When $$$i = 1$$$ is chosen, $$$x = |2-3| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

In the third sample, $$$a = [3, 4, 5, 6, 7]$$$. When $$$i = 3$$$ is chosen, $$$x = |3 + 4 + 5 - 6 - 7| = 1$$$. It can be shown this is the minimum possible value of $$$x$$$.

Buggy Code:
import java.util.*;
public class test {

    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();

        while (t-- > 0) {
            long n = sc.nextInt();
             long k=sc.nextInt();
             if(n==1000000000 && k==1000000000)
             {
                System.out.println(347369930);
                continue;
             }
            long total = (long) n * (2*k +(n-1)) / 2;
            long min=Integer.MAX_VALUE;
            long left=0;
            for(int i=0;i<n;i++)
            {
                left+=k+i;
                long right=total-left;
                if(Math.abs(left-right)<min)
                {
                    min=(long) Math.abs(left-right);
                }
            }
            System.out.println(min);
    }
}
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
1
1000000000 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

Input: Each test contains multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int maxRemainingElement(vector<int>& a) {
    int n = a.size();
    if (n == 1) return a[0];

    int maxElement = 0;
    for (int i = 0; i < n - 1; ++i) {
        vector<int> newA;
        for (int j = 0; j < n; ++j) {
            if (j != i && j != i + 1) {
                newA.push_back(a[j]);
            }
        }
        maxElement = max(maxElement, maxRemainingElement(newA));
    }
    return maxElement;
}

int main() {
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        vector<int> a(n);
        for (int i = 0; i < n; i++) {
            cin >> a[i];
        }
        cout << maxRemainingElement(a) << endl;
    }
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
1
100
3
100 1 100
5
100 1 1 1 100
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

Input: Each test contains multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int t;
    cin >> t;
    while(t--){
        int n;
        cin>>n;
        int arr[n];
        for(int i=0;i<n;i++){
            cin>>arr[i];
        }
        int ans = arr[n-1]>arr[n-3]? arr[n-1]:arr[n-3];
        cout<<ans<<endl;
    }
    return 0;
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to create a test case where the code's logic of comparing only the last and third-to-last elements will fail to find the maximum possible remaining element.

The buggy code only compares the last and third-to-last elements:
```cpp
int ans = arr[n-1] > arr[n-3] ? arr[n-1] : arr[n-3];
```

This logic will fail when the maximum element is not in one of these positions. Let's create a test case where the maximum element is in the middle of the array, and the last and third-to-last elements are smaller.

### Test Case:
```plaintext
1
5
1 5 3 2 4
```

### Explanation:
- The array is `[1, 5, 3, 2, 4]`.
- The last element is `4`.
- The third-to-last element is `3`.
- The maximum element in the array is `5`.

The buggy code will output `4` (since `4 > 3`), but the correct answer should be `5`.

### Output:
```plaintext
5
```

This test case will expose the bug in the buggy code because the maximum element is not in the last or third-to-last position.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

Input: Each test contains multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
int main()
{
    int t, n;
    cin >> t;
    while (t--)
    {
        cin >> n;
        vector<int> vect;
        int k;
        for (int i = 0; i < n; i++)
        {
            cin >> k;
            vect.push_back(k);
        }
        int maxi = 0;
        int index = 0;
        int pre = 0;
        int index2 = 0;
        for (int i = 0; i < n; i++)
        {
            if (maxi < vect[i])
            {
                pre = maxi;
                maxi = vect[i];

                index = i;
            }
            else if (pre < vect[i] && vect[i] < maxi)
            {
                pre = vect[i];
                index2 = i;
            }
        }
        if (index != n - 2 && index != 1)
        {
            cout << maxi;
        }
        else
        {
            maxi = 0;
            if (index2 != n - 2 && index2 != 1)
                cout << pre;

            else
            {
                for (int i = 2; i < n - 2; i++)
                {
                    if (maxi < vect[i])
                        maxi = vect[i];
                }
                if (maxi > vect[0] && maxi > vect[n - 1])
                    cout << maxi;
                else if (vect[0] > vect[n - 1])
                    cout << vect[0];
                else
                    cout << vect[n - 1];
            }
        }
        cout << "\n";
    }
}

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
3
5
1 2 3 2 1
7
1 2 3 4 3 2 1
9
1 2 3 4 5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Maximize the Last Element
time_limit_ms: 1000
memory_limit_mb: 256
Description: You are given an array $a$ of $n$ integers, where $n$ is odd.

In one operation, you will remove two adjacent elements from the array $a$, and then concatenate the remaining parts of the array. For example, given the array $[4,7,4,2,9]$, we can obtain the arrays $[4,2,9]$ and $[4,7,9]$ by the operations $[\underline{4,7}, 4,2,9] \to [4,2,9]$ and $[4,7,\underline{4,2},9] \to [4,7,9]$ respectively. However, we cannot obtain the array $[7,2,9]$ as it requires deleting non-adjacent elements $[\underline{4},7,\underline{4},2,9]$.

You will repeatedly perform this operation until exactly one element remains in $a$.

Find the maximum possible value of the remaining element in $a$.

Input: Each test contains multiple test cases. The first line contains a single integer $t$ ($1 \le t \le 1000$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($1 \le n \le 99$; $n$ is odd) — the length of the array $a$.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 100$) — the elements of the array $a$.

Note that there is no bound on the sum of $n$ over all test cases.

Output: For each test case, output a single integer — the maximum possible value of the remaining element in $a$.

Examples:
input:
4
1
6
3
1 3 2
5
4 7 4 2 9
7
3 1 4 1 5 9 2
output:
6
2
9
5

Note: In the first test case, the array $a$ is $[6]$. Since there is only one element, no operations are needed. The maximum possible value of the remaining element is $6$.

In the second test case, the array $a$ is $[1, 3, 2]$. We can remove the first two elements $[\underline{1, 3}, 2] \to [2]$, or remove the last two elements $[1, \underline{3, 2}] \to [1]$. Therefore, the maximum possible value of the remaining element is $2$.

In the third test case, the array $a$ is $[4, 7, 4, 2, 9]$. One way to maximize the remaining element is $[4, \underline{7, 4}, 2, 9] \to [\underline{4, 2}, 9] \to [9]$. Therefore, the maximum possible value of the remaining element is $9$.

In the fourth test case, the array $a$ is $[3, 1, 4, 1, 5, 9, 2]$. It can be shown that the maximum possible value of the remaining element is $5$.

Buggy Code:
t = int(input())
for _ in range(t):
    n = int(input())
    a = list(map(int, input().split(" ")))
    first_max = 0
    second_max = 0
    for i in range(1,len(a)):
        if a[i] > a[first_max]:
            second_max = first_max
            first_max = i
        elif a[i] > a[second_max]:
            second_max = i
    count = len(a)-first_max-1

    if count%2 == 0:
        print(a[first_max])
    else:
        print(a[second_max])

**Instructions:**
- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.
- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.
- Output the test input directly, not code to generate it.

**Output format:**
```plaintext
{test input}
```

Think step by step.

Response:
```plaintext
4
1
100
3
100 1 100
5
100 1 100 1 100
7
1 100 1 100 1 100 1
```