**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

Buggy Code:
import java.lang.reflect.Array;
import java.math.BigInteger;
import java.util.*;
import java.io.*;

public class Main {

    public static int difference (int[] a, int position, int k)
    {
        int minimum = a[position];
        int index = -1;
        for (int i = position+1; i < a.length && i <= position + k; i++)
        {
            if (a[i] < minimum)
            {
                minimum = a[i];
                index = i;
            }
        }
        for (int i = position-1; i >= 0 && i >= position - k; i--)
        {
            if (a[i] < minimum)
            {
                minimum = a[i];
                index = i;
            }
        }
        return index;
    }

    public static void main(String[] args) throws IOException{
        FastIO io = new FastIO();
        int testcases = io.nextInt();
        outer:
        for (int t = 0; t < testcases; t++) {
            int n;
            int k;
            int[] a = new int [n = io.nextInt()];
            k = io.nextInt();
            for (int i = 0; i < n ;i++)
                a[i] = io.nextInt();
            if (n == 1)
            {
                io.println(a[0]);
                continue ;
            }
            while (k > 0)
            {
                int start = -1;
                int finish = -1;
                int bestDiff = 0;
                for (int i = 0; i < n; i++)
                {
                    int index = difference(a, i, k);
                    if (index == -1)
                        continue;
                    if (a[i] - a[index] > bestDiff)
                    {
                        bestDiff = a[i] - a[index];
                        start = index;
                        finish = i;
                    }
                }
                if (start == -1)
                    break;
                int changeTo = Math.min(a[start], a[finish]);
                for (int i = Math.min(start, finish); i <= Math.max(start, finish); i++)
                    a[i] = changeTo;
                k -= Math.abs(finish - start);
            }
            long sum = 0;
            for (int i = 0; i < n; i++)
                sum += a[i];
            io.println(sum);
        }
        io.flush();
    }

    static class FastIO extends PrintWriter {
        private InputStream stream;
        private byte[] buf = new byte[1 << 16];
        private int curChar, numChars;
        // standard input
        public FastIO() {
            this(System.in, System.out);
        }
        public FastIO(InputStream i, OutputStream o) {
            super(o);
            stream = i;
        }
        // file input
        public FastIO(String i, String o) throws IOException {
            super(new FileWriter(o));
            stream = new FileInputStream(i);
        }
        // throws InputMismatchException() if previously detected end of file
        private int nextByte() {
            if (numChars == -1) throw new InputMismatchException();
            if (curChar >= numChars) {
                curChar = 0;
                try {
                    numChars = stream.read(buf);
                } catch (IOException e) {
                    throw new InputMismatchException();
                }
                if (numChars == -1) return -1;  // end of file
            }
            return buf[curChar++];
        }
        // to read in entire lines, replace c <= ' '
        // with a function that checks whether c is a line break
        public String next() {
            int c;
            do {
                c = nextByte();
            } while (c <= ' ');
            StringBuilder res = new StringBuilder();
            do {
                res.appendCodePoint(c);
                c = nextByte();
            } while (c > ' ');
            return res.toString();
        }
        public int nextInt() {  // nextLong() would be implemented similarly
            int c;
            do {
                c = nextByte();
            } while (c <= ' ');
            int sgn = 1;
            if (c == '-') {
                sgn = -1;
                c = nextByte();
            }
            int res = 0;
            do {
                if (c < '0' || c > '9') throw new InputMismatchException();
                res = 10 * res + c - '0';
                c = nextByte();
            } while (c > ' ');
            return res * sgn;
        }
        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }
    public static class SegmentTree
    {
        private long[] lazy, A, st;
        private long neutral = 0;
        private int n, size;
        public SegmentTree (long[] B)
        {
            int size1 = B.length;
            n = 1;
            while (n < size1)
                n <<= 1;
            size1 = n;
            long[] trueB = new long [size1];
            for (int i = 0; i < B.length; i++)
                trueB[i] = B[i];
            A = trueB;
            n <<= 1;
            st = new long [n];
            lazy = new long [n];
            size = size1;
            Arrays.fill(st, neutral);
            Arrays.fill(lazy, 0);
            build(1,0,size-1);
        }

        public static long merge (long val1, long val2)
        {
            return val1 + val2;
        }
        public void build (int index, int l, int r)
        {
            if (l == r)
            {
                st[index] = A[l];
            }
            else
            {
                int mid = (l+r) >> 1;
                build (index << 1, l, mid);
                build ((index << 1) | 1, mid+1, r);
                st[index] = merge(st[index << 1], st[(index << 1) | 1]);
            }
        }

        public void update (int l, int r, long value)
        {
            update (0, size-1, 1, l, r, value);
        }

        private void update (int start, int end, int index, int l, int r, long value)
        {
            if (l > end || r < start)
                return;
            else if (start == end)
            {
                st[start] += value;
            }
            else
            {
                int mid = (start + end) >> 1;
                update(start, mid, index << 1, l, r, value);
                update(mid+1, end, (index << 1) | 1, l, r, value);
                st[index] = merge(st[index << 1], st[(index<<1)|1]);
            }
        }

        public void propagate (int index)
        {
            int left = index<<1;
            int right = (index << 1) | 1;
            if (index >= size)
            {
                if (lazy[index] == 0)
                    return;
                st[index] += lazy[index];
                lazy[index] = 0;
                return;
            }
            if (lazy[index] != 0)
            {
                lazy[left] = lazy[left] == 0 ? lazy[index]/2 : lazy[left] + lazy[index]/2;
                lazy[right] = lazy[right] == 0 ? lazy[index]/2 : lazy[right] + lazy[index]/2;
                st[index] += lazy[index];
                lazy[index] = 0;
            }
        }

        public long query (int i, int j)
        {
            return query (0, size-1, 1, i, j);
        }

        private long query (int start, int end, int index, int l, int r)
        {
            if (start == end)
            {
                return st[index];
            }
            else if (l > end || r < start)
                return neutral;
            else
            {
                int mid = (start + end) >> 1;
                long s1 = query(start, mid, index << 1, l, r);
                long s2 = query(mid+1, end, (index << 1) | 1, l, r);
                return merge(s1, s2);
            }
        }
    }

    public static int LSOne(int S)
    {
        return ( (S) & -(S) );
    }

    static class SegmentTreeSet {                              // OOP style
        private int n;                                         // n = (int)A.size()
        private long[] A, st, lazy;                                // the arrays

        private int l(int p) { return  p<<1; }                 // go to left child
        private int r(int p) { return (p<<1)+1; }              // go to right child

        private long conquer(long a, long b) {
            if (a == -1) return b;                       // corner case
            if (b == -1) return a;
            return a+b;                            // RMQ
        }

        private void build(int p, int L, int R) {              // O(n)
            if (L == R)
                st[p] = A[L];                              // base case
            else {
                int m = (L+R)/2;
                build(l(p), L  , m);
                build(r(p), m+1, R);
                st[p] = conquer(st[l(p)], st[r(p)]);
            }
        }

        private void propagate(int p, int L, int R) {
            if (lazy[p] != -1) {                         // has a lazy flag
                st[p] = lazy[p];                           // [L..R] has same value
                if (L != R)                                // not a leaf
                    lazy[l(p)] = lazy[r(p)] = lazy[p];       // propagate downwards
                else                                       // L == R, a single index
                    A[L] = lazy[p];                          // time to update this
                lazy[p] = -1;                              // erase lazy flag
            }
        }

        private long RMQ(int p, int L, int R, int i, int j) {   // O(log n)
            propagate(p, L, R);                          // lazy propagation
            if (i > j) return -1;                        // infeasible
            if ((L >= i) && (R <= j)) return st[p];      // found the segment
            int m = (L+R)/2;
            return conquer(RMQ(l(p), L, m, i, Math.min(m, j)),
                    RMQ(r(p), m+1, R, Math.max(i, m+1), j));
        }

        private void update(int p, int L, int R, int i, int j, int val) { // O(log n)
            propagate(p, L, R);                          // lazy propagation
            if (i > j) return;
            if ((L >= i) && (R <= j)) {                  // found the segment
                lazy[p] = val;                             // update this
                propagate(p, L, R);                        // lazy propagation
            }
            else {
                int m = (L+R)/2;
                update(l(p), L  , m, i, Math.min(m, j), val);
                update(r(p), m+1, R, Math.max(i, m+1), j        , val);
                st[p] = conquer(st[l(p)], st[r(p)]);
            }
        }

        public SegmentTreeSet(int sz)
        {
            n = sz;
            st = new long [4*n];
            lazy = new long [4*n];
            for (int i = 0 ; i < 4*n; i++)
                lazy[i] = -1;
        }

        public SegmentTreeSet(long[] initialA)
        {
            this(initialA.length);
            A = initialA;
            build(1, 0, n-1);
        }

        public void update(int i, int j, int val) { update(1, 0, n-1, i, j, val); }

        public long RMQ(int i, int j) { return RMQ(1, 0, n-1, i, j); }

        int getMid(int s, int e) {
            return s + (e - s) / 2;
        }
    };

    static class FenwickTree {
        int[] ft;
        public FenwickTree(int m)
        {
            ft = new int[m+1];
        }
        public void build (int[] f)
        {
            int m = f.length -1;
            ft = new int [m+1];
            for (int i = 0; i < m+1 ;i++)
                ft[i] = 0;
            for (int i = 1; i <= m; ++i)
            {
                ft[i] += f[i];
                if (i+LSOne(i) <= m)
                    ft[i+LSOne(i)] += ft[i];
            }
        }

        FenwickTree (int[] f)
        {
            build(f);
        }

        FenwickTree(int m, int[] s)
        {
            int[] f = new int [m+1];
            for (int i = 0; i < s.length; i++)
            {
                ++f[s[i]];
            }
            build(f);
        }

        long rsq (int j)
        {
            long sum = 0;
            for (; j > 0; j -= LSOne(j))
                sum += ft[j];
            return sum;
        }

        long rsq (int i, int j) {return rsq(j) - rsq(i-1);}

        //updates value of the i-th element by v (v can be +ve/inc or -ve/dec)
        public void update(int i, long v)
        {
            while (i < ft.length) {
                ft[i] += v;
                i += LSOne(i);
            }
        }

        int select (int k) {
            int lo = 1, hi = ft.length-1;
            for (int i =0; i < 30; ++i)
            {
                int mid = (lo+hi)/2;
                if (rsq(1, mid) < k)
                    lo = mid;
                else
                    hi = mid;
            }
            return hi;
        }
    };

    static class RUPQ {                                     // RUPQ variant
        private FenwickTree ft;                                // internally use PURQ FT
        public RUPQ(int m)
        {
            ft = new FenwickTree(m);
        }
        void range_update(int ui, int uj, long v) {
            ft.update(ui, v);                            // [ui, ui+1, .., m] +v
            ft.update(uj+1, -v);                         // [uj+1, uj+2, .., m] -v
        }                                              // [ui, ui+1, .., uj] +v
        long point_query(int i) { return ft.rsq(i); }    // rsq(i) is sufficient
    };

    static class RURQ  {                                    // RURQ variant
        private RUPQ rupq;                                     // one RUPQ and
        private FenwickTree purq;                              // one PURQ
        public RURQ(int m){
            rupq = new RUPQ(m);
            purq = new FenwickTree(m);
        }
        public void range_update(int ui, int uj, long v) {
            rupq.range_update(ui, uj, v);                // [ui, ui+1, .., uj] +v
            purq.update(ui, v*(ui-1));                   // -(ui-1)*v before ui
            purq.update(uj+1, -v*uj);                    // +(uj-ui+1)*v after uj
        }
        public long rsq(int j) {
            return rupq.point_query(j)*j -               // optimistic calculation
                    purq.rsq(j);                          // cancelation factor
        }
        public long rsq(int i, int j) { return rsq(j) - rsq(i-1); } // standard
    };

    static class Triplet implements Comparable{
        int a;
        int b;
        int c;

        public Triplet(int a, int b, int c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }

        public int compareTo (Object o)
        {
            Triplet t= (Triplet) o;
            if (b != t.b)
                return b - t.b;
            if (a != t.a)
                return a - t.a;
            return c - t.c;
            //           return a == t.a ? b == t.b ? c - t.c : b - t.b : a - t.a;
        }
        public String toString() {
            return String.valueOf(this.a) + " " + this.b + " " + this.c;
        }

    }

    public static int digSum(int n) {
        if (n == 0)
            return 0;
        return (n % 9 == 0) ? 9 : (n % 9);
    }

    public static boolean[] sieveOfEratosthenes(int n) {
        // Create a boolean array "prime[0..n]" and initialize
        // all entries it as true. A value in prime[i] will
        // finally be false if i is Not a prime, else true.
        boolean prime[] = new boolean[n + 1];
        for (int i = 0; i <= n; i++)
            prime[i] = true;

        for (int p = 2; p * p <= n; p++) {
            // If prime[p] is not changed, then it is a prime
            if (prime[p] == true) {
                // Update all multiples of p
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }
        return prime;
    }

    static long gcd(long a, long b) {
        if (a == 0)
            return b;
        return gcd(b % a, a);
    }

    // method to return LCM of two numbers
    static long lcm(long a, long b) {
        return (a / gcd(a, b)) * b;
    }

    static class COUNT {
        long a;

        public COUNT(long a) {
            this.a = a;
        }

        public void add() {
            this.a++;
        }

        public void add(int value) {
            this.a = this.a + value;
        }

        public void subtract() {
            this.a--;
        }

        public String toString() {
            return String.valueOf(a);
        }

    }

    static boolean isPrime(long num) {
        if (num == 1)
            return false;
        if (num <= 3)
            return true;
        if (num % 2 == 0 || num % 3 == 0 || num % (int) Math.sqrt(num) == 0)
            return false;
        for (int i = 4; i < (int) Math.sqrt(num); i++) {
            if (num % i == 0)
                return false;
        }
        return true;
    }

    public static class Node {
        public boolean next;
        int node;
        int edge;

        public Node(int node, int edge) {
            this.node = node;
            this.edge = edge;
        }

        public String toString()
        {
            return this.node + " " + this.edge + "\n";
        }

    }

    static class MIN {
        long a;

        public MIN() {
            this.a = Integer.MAX_VALUE;
        }

        public void add(long a) {
            if (a < this.a)
                this.a = a;
        }

        public String toString() {
            return String.valueOf(this.a);
        }
    }

    static class MAX {
        long a;

        public MAX() {
            this.a = -1;
        }

        public void add(long a) {
            if (a > this.a)
                this.a = a;
        }

        public String toString() {
            return String.valueOf(this.a);
        }

    }

    static class Pair implements Comparable<Pair>{
        int a,b;
        public Pair (int a, int b)
        {
            this.a = a;
            this.b = b;
        }
        public String toString() {
            return "" + this.a + " | " + this.b;
        }

        @Override
        public int hashCode() {
            return Objects.hash(a, b);
        }

        @Override
        public boolean equals(Object a1) {
            if (this == a1)
                return true;
            if (a1 == null || getClass() != a1.getClass())
                return false;
            Pair idk = (Pair) a1;
            return a == idk.a && b == idk.b;
        }

        public int compareTo(Pair u1)
        {
            if (u1.a == a)
                return Integer.compare(b, u1.b);
            return Integer.compare(a, u1.a);
        }
    }

    private static void helper(List<int[]> combinations, int data[], int start, int end, int index) {
        if (index == data.length) {
            int[] combination = data.clone();
            combinations.add(combination);
        } else if (start <= end) {
            data[index] = start;
            helper(combinations, data, start + 1, end, index + 1);
            helper(combinations, data, start + 1, end, index);
        }
    }

    public static List<int[]> generate(int n, int r) {
        List<int[]> combinations = new ArrayList<>();
        helper(combinations, new int[r], 0, n - 1, 0);
        return combinations;
    }

    public static double round_to_d_decimal_places(double n, int d) {
        n *= Math.pow(10, d);
        n = Math.round(n);
        n /= Math.pow(10, d);
        return n;
    }

    static class UnionFind {
        private ArrayList<Integer> p, rank, setSize;
        private int numSets;

        public UnionFind(int N) {
            p = new ArrayList<>(N);
            rank = new ArrayList<>(N);
            setSize = new ArrayList<>(N);
            numSets = N;
            for (int i = 0; i < N; i++) {
                p.add(i);
                rank.add(0);
                setSize.add(1);
            }
        }

        public int findSet(int i) {
            if (p.get(i) == i) return i;
            else {
                int ret = findSet(p.get(i)); p.set(i, ret);
                return ret; } }

        public Boolean isSameSet(int i, int j) { return findSet(i) == findSet(j); }

        public void unionSet(int i, int j) {
            if (!isSameSet(i, j)) { numSets--;
                int x = findSet(i), y = findSet(j);
                // rank is used to keep the tree short
                if (rank.get(x) > rank.get(y)) { p.set(y, x); setSize.set(x, setSize.get(x) + setSize.get(y)); }
                else                           { p.set(x, y); setSize.set(y, setSize.get(y) + setSize.get(x));
                    if (rank.get(x) == rank.get(y)) rank.set(y, rank.get(y) + 1); } } }
        public int numDisjointSets() { return numSets; }
        public int sizeOfSet(int i) { return setSize.get(findSet(i)); }
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

        public Scanner(InputStream s) {
            br = new BufferedReader(new InputStreamReader(s));
        }

        public Scanner(String file) throws IOException {
            br = new BufferedReader(new FileReader(file));
        }

        public Scanner(FileReader r) {
            br = new BufferedReader(r);
        }

        public String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }

        public String readAllLines(BufferedReader reader) throws IOException {
            StringBuilder content = new StringBuilder();
            String line;
            while ((line = reader.readLine()) != null) {
                content.append(line);
                content.append(System.lineSeparator());
            }
            return content.toString();
        }

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

        public long nextLong() throws IOException {
            return Long.parseLong(next());
        }

        public String nextLine() throws IOException {
            return br.readLine();
        }

        public double nextDouble() throws IOException {
            String x = next();
            StringBuilder sb = new StringBuilder("0");
            double res = 0, f = 1;
            boolean dec = false, neg = false;
            int start = 0;
            if (x.charAt(0) == '-') {
                neg = true;
                start++;
            }
            for (int i = start; i < x.length(); i++)
                if (x.charAt(i) == '.') {
                    res = Long.parseLong(sb.toString());
                    sb = new StringBuilder("0");
                    dec = true;
                } else {
                    sb.append(x.charAt(i));
                    if (dec)
                        f *= 10;
                }
            res += Long.parseLong(sb.toString()) / f;
            return res * (neg ? -1 : 1);
        }

        public long[] nextlongArray(int n) throws IOException {
            long[] a = new long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public Long[] nextLongArray(int n) throws IOException {
            Long[] a = new Long[n];
            for (int i = 0; i < n; i++)
                a[i] = nextLong();
            return a;
        }

        public int[] nextIntArray(int n) throws IOException {
            int[] a = new int[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public Integer[] nextIntegerArray(int n) throws IOException {
            Integer[] a = new Integer[n];
            for (int i = 0; i < n; i++)
                a[i] = nextInt();
            return a;
        }

        public boolean ready() throws IOException {
            return br.ready();
        }
    }

    public static class RedBlackTree {

        class Node {
            int data;
            Node parent;
            Node left;
            Node right;
            int color;
            int rank;  // New field to represent the rank

            public Node() {
                this.rank = 0;  // Initialize the rank to 0
            }
        }

        private Node root;
        private Node TNULL;

        // Preorder
        private void preOrderHelper(Node node) {
            if (node != TNULL) {
                System.out.print(node.data + " ");
                preOrderHelper(node.left);
                preOrderHelper(node.right);
            }
        }

        // Inorder
        private void inOrderHelper(Node node) {
            if (node != TNULL) {
                inOrderHelper(node.left);
                System.out.print(node.data + " ");
                inOrderHelper(node.right);
            }
        }

        // Post order
        private void postOrderHelper(Node node) {
            if (node != TNULL) {
                postOrderHelper(node.left);
                postOrderHelper(node.right);
                System.out.print(node.data + " ");
            }
        }

        // Search the tree
        private Node searchTreeHelper(Node node, int key) {
            if (node == TNULL || key == node.data) {
                return node;
            }

            if (key < node.data) {
                return searchTreeHelper(node.left, key);
            }
            return searchTreeHelper(node.right, key);
        }

        private void rbTransplant(Node u, Node v) {
            if (u.parent == null) {
                root = v;
            } else if (u == u.parent.left) {
                u.parent.left = v;
            } else {
                u.parent.right = v;
            }
            if (v != null) {
                v.parent = u.parent;  // Make sure to update the parent of v
            }
        }

        private void deleteNodeHelper(Node node, int key) {
            Node z = TNULL;
            Node x, y;
            while (node != TNULL) {
                if (node.data == key) {
                    z = node;
                }

                if (node.data <= key) {
                    node = node.right;
                } else {
                    node = node.left;
                }
            }

            if (z == TNULL) {
                System.out.println("Couldn't find key in the tree");
                return;
            }

            y = z;
            int yOriginalColor = y.color;
            if (z.left == TNULL) {
                x = z.right;
                rbTransplant(z, z.right);
            } else if (z.right == TNULL) {
                x = z.left;
                rbTransplant(z, z.left);
            } else {
                y = minimum(z.right);
                yOriginalColor = y.color;
                x = y.right;
                if (y.parent == z) {
                    x.parent = y;
                } else {
                    rbTransplant(y, y.right);
                    y.right = z.right;
                    y.right.parent = y;
                }

                rbTransplant(z, y);
                y.left = z.left;
                y.left.parent = y;
                y.color = z.color;
            }
            if (yOriginalColor == 0) {
                fixDelete(x);
            }
        }

        private void fixInsert(Node k) {
            Node u;
            while (k.parent != null && k.parent.color == 1) {
                if (k.parent == k.parent.parent.right) {
                    u = k.parent.parent.left;
                    if (u != null && u.color == 1) {
                        u.color = 0;
                        k.parent.color = 0;
                        k.parent.parent.color = 1;
                        k = k.parent.parent;
                    } else {
                        if (k == k.parent.left) {
                            k = k.parent;
                            rightRotate(k);
                        }
                        if (k.parent != null) {
                            k.parent.color = 0;
                            if (k.parent.parent != null) {
                                k.parent.parent.color = 1;
                                leftRotate(k.parent.parent);
                            }
                        }
                    }
                } else {
                    u = k.parent.parent.right;
                    if (u != null && u.color == 1) {
                        u.color = 0;
                        k.parent.color = 0;
                        k.parent.parent.color = 1;
                        k = k.parent.parent;
                    } else {
                        if (k == k.parent.right) {
                            k = k.parent;
                            leftRotate(k);
                        }
                        if (k.parent != null) {
                            k.parent.color = 0;
                            if (k.parent.parent != null) {
                                k.parent.parent.color = 1;
                                rightRotate(k.parent.parent);
                            }
                        }
                    }
                }
                if (k == root) {
                    break;
                }
            }
            root.color = 0;
        }

        private void leftRotate(Node x) {
            Node y = x.right;
            x.right = y.left;
            if (y.left != TNULL) {
                y.left.parent = x;
            }
            y.parent = x.parent;
            if (x.parent == null) {
                this.root = y;
            } else if (x == x.parent.left) {
                x.parent.left = y;
            } else {
                x.parent.right = y;
            }
            y.left = x;
            x.parent = y;
        }

        // Delete a node with the given key
        public void deleteNode(int key) {
            deleteNodeHelper(this.root, key);
        }

        private void rightRotate(Node x) {
            Node y = x.left;
            x.left = y.right;
            if (y.right != TNULL) {
                y.right.parent = x;
            }
            y.parent = x.parent;
            if (x.parent == null) {
                this.root = y;
            } else if (x == x.parent.right) {
                x.parent.right = y;
            } else {
                x.parent.left = y;
            }
            y.right = x;
            x.parent = y;
        }

        public void insert(int key) {
            Node node = new Node();
            node.parent = null;
            node.data = key;
            node.left = TNULL;
            node.right = TNULL;
            node.color = 1;  // new node must be red

            Node y = null;
            Node x = this.root;

            while (x != TNULL) {
                y = x;
                if (node.data < x.data) {
                    x.rank++;  // Increase rank when moving left
                    x = x.left;
                } else {
                    node.rank = y.rank + 1;  // Set rank to parent's rank + 1 when moving right
                    x = x.right;
                }
            }

            node.parent = y;
            if (y == null) {
                root = node;
            } else if (node.data < y.data) {
                y.left = node;
            } else {
                y.right = node;
            }

            if (node.parent != null) {
                fixInsert(node);
            } else {
                root.color = 0;
            }
        }

        // Find the rank of a given key in the sorted order
        public int findRank(int key) {
            return findRankHelper(root, key);
        }

        private int findRankHelper(Node node, int key) {
            if (node == TNULL) {
                return 0;  // Element not found
            }

            if (key < node.data) {
                return findRankHelper(node.left, key);
            } else if (key > node.data) {
                return node.rank + findRankHelper(node.right, key);
            } else {
                return node.rank;  // Element found, return its rank
            }
        }

        // Find the rank of a given key in the sorted order

        // Helper method to get the size of the subtree rooted at a given node
        private int size(Node node) {
            if (node == TNULL) {
                return 0;
            }
            return 1 + size(node.left) + size(node.right);
        }

        // Balancing the tree after deletion of a node
        private void fixDelete(Node x) {
            Node s;
            while (x != root && x.color == 0) {
                if (x == x.parent.left) {
                    s = x.parent.right;
                    if (s.color == 1) {
                        s.color = 0;
                        x.parent.color = 1;
                        leftRotate(x.parent);
                        s = x.parent.right;
                    }

                    if (s.left.color == 0 && s.right.color == 0) {
                        s.color = 1;
                        x = x.parent;
                    } else {
                        if (s.right.color == 0) {
                            s.left.color = 0;
                            s.color = 1;
                            rightRotate(s);
                            s = x.parent.right;
                        }

                        s.color = x.parent.color;
                        x.parent.color = 0;
                        s.right.color = 0;
                        leftRotate(x.parent);
                        x = root;
                    }
                } else {
                    s = x.parent.left;
                    if (s.color == 1) {
                        s.color = 0;
                        x.parent.color = 1;
                        rightRotate(x.parent);
                        s = x.parent.left;
                    }

                    if (s.right.color == 0 && s.right.color == 0) {
                        s.color = 1;
                        x = x.parent;
                    } else {
                        if (s.left.color == 0) {
                            s.right.color = 0;
                            s.color = 1;
                            leftRotate(s);
                            s = x.parent.left;
                        }

                        s.color = x.parent.color;
                        x.parent.color = 0;
                        s.left.color = 0;
                        rightRotate(x.parent);
                        x = root;
                    }
                }
            }
            x.color = 0;
        }

        public int size() {
            return size(root);
        }

        // Finding the minimum key value node in a subtree
        private Node minimum(Node node) {
            while (node.left != TNULL) {
                node = node.left;
            }
            return node;
        }

        // Print the tree
        public void printTree() {
            inOrderHelper(this.root);
            System.out.println();
        }
    }
}
//0 1 0 1 0 1 0 1 0 1

**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 test input that exposes the bug in the given buggy code, we need to focus on scenarios that are likely to trigger the faulty logic. The main issue in the code seems to be related to the `difference` function and the way it handles the indices and the `k` operations.

### Step-by-Step Reasoning:
1. **Edge Cases**:
   - **Single Element Array**: The code should handle this case correctly, but it's a good starting point.
   - **Array with Repeated Elements**: This can test the logic for choosing the best element to replace.
   - **Array with Elements in Increasing/Decreasing Order**: This can test the logic for finding the minimum element within the allowed range.
   - **Array with Elements Having Large Differences**: This can test the logic for maximizing the reduction in the sum.

2. **Small-Scale Test Input**:
   - We will create a small array with a few elements and a small `k` value to ensure the test input is manageable and easy to verify.

### Test Input:
Let's create an array with elements that have significant differences and a small `k` value to see if the code can correctly minimize the sum.

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

### Explanation:
- **Test Case 1**:
  - **Array**: `[5, 1, 4]`
  - **k**: `2`
  - **Expected Output**: `6`
  - **Reasoning**:
    - In the first operation, we can replace `5` with `1` to get `[1, 1, 4]`.
    - In the second operation, we can replace `4` with `1` to get `[1, 1, 1]`.
    - The total sum is `1 + 1 + 1 = 3`.

This test input should expose the bug in the `difference` function and the logic for choosing the best element to replace. The expected output is `6`, but the correct output should be `3` if the code is working correctly.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Minimizing the Sum
time_limit_ms: 2000
memory_limit_mb: 256
Description: You are given an integer array $$$a$$$ of length $$$n$$$.

You can perform the following operation: choose an element of the array and replace it with any of its neighbor's value.

For example, if $$$a=[3, 1, 2]$$$, you can get one of the arrays $$$[3, 3, 2]$$$, $$$[3, 2, 2]$$$ and $$$[1, 1, 2]$$$ using one operation, but not $$$[2, 1, 2$$$] or $$$[3, 4, 2]$$$.

Your task is to calculate the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

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

The first line of each test case contains two integers $$$n$$$ and $$$k$$$ ($$$1 \le n \le 3 \cdot 10^5$$$; $$$0 \le k \le 10$$$).

The second line contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$1 \le a_i \le 10^9$$$).

Additional constraint on the input: the sum of $$$n$$$ over all test cases doesn't exceed $$$3 \cdot 10^5$$$.

Output: For each test case, print a single integer — the minimum possible total sum of the array if you can perform the aforementioned operation at most $$$k$$$ times.

Examples:
input:
4
3 1
3 1 2
1 3
5
4 2
2 2 1 3
6 3
4 1 2 2 4 3
output:
4
5
5
10

Note: In the first example, one of the possible sequences of operations is the following: $$$[3, 1, 2] \rightarrow [1, 1, 2$$$].

In the second example, you do not need to apply the operation.

In the third example, one of the possible sequences of operations is the following: $$$[2, 2, 1, 3] \rightarrow [2, 1, 1, 3] \rightarrow [2, 1, 1, 1]$$$.

In the fourth example, one of the possible sequences of operations is the following: $$$[4, 1, 2, 2, 4, 3] \rightarrow [1, 1, 2, 2, 4, 3] \rightarrow [1, 1, 1, 2, 4, 3] \rightarrow [1, 1, 1, 2, 2, 3]$$$.

Buggy Code:
import java.util.Arrays;
import java.util.Scanner;

public class one {

    static long[][][][] dp;

    // Recursive function with memoization
    static long rec(int i, int m, int cs, int b, long[] arr) {
        int n = arr.length;

        // Base case: If we've reached the end of the array
        if (i == n) {
            return 0; // If operations are left, it's invalid
        }

        // Return cached result if available
        if (dp[i][m][cs][b] != -1) {
            return dp[i][m][cs][b];
        }

        long ans = Long.MAX_VALUE;

        // Check all valid transitions
        if (m > 0) {
            if (b == 0) { // Last operation was front
                if (i > 0) { // Back operation possible
                    ans = Math.min(ans, rec(i + 1, m - 1, 1, 2, arr) + arr[i - 1]);
                }
                if (i + 1 < n) { // Front operation possible
                    ans = Math.min(ans, rec(i + 1, m - 1, 0, 1, arr) + arr[i + 1]);
                }
            } else if (b == 2) { // Last operation was back
                if (i - cs - 1 >= 0) { // Continue back operations
                    ans = Math.min(ans, rec(i + 1, m - 1, cs + 1, 2, arr) + arr[i - cs - 1]);
                }
                if (i + 1 < n) { // Front operation after back
                    ans = Math.min(ans, rec(i + 1, m - 1, 0, 1, arr) + arr[i + 1]);
                }
            } else if (b == 1) { // Last operation was front
                if (i + 1 < n) { // Front operation
                    ans = Math.min(ans, rec(i + 1, m - 1, 0, 1, arr) + arr[i + 1]);
                }
            }
        }

        // No operation, just move to the next index
        ans = Math.min(ans, rec(i + 1, m, 0, 0, arr) + arr[i]);

        // Store and return result
        return dp[i][m][cs][b] = ans;
    }

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

        while (t-- > 0) {
            int n = sc.nextInt();
            int k = sc.nextInt();
            long[] arr = new long[n];
            for (int i = 0; i < n; i++) {
                arr[i] = sc.nextLong();
            }

            // Initialize DP array
            dp = new long[n + 1][k + 1][k + 1][3];
            for (long[][][] table : dp) {
                for (long[][] row : table) {
                    for (long[] col : row) {
                        Arrays.fill(col, -1);
                    }
                }
            }

            // Solve for the original array
            long ans = rec(0, k, 0, 0, arr);

            // Solve for the reversed array
            long[] reversedArr = new long[n];
            for (int i = 0; i < n; i++) {
                reversedArr[i] = arr[n - 1 - i];
            }

            // Reinitialize DP array
            dp = new long[n + 1][k + 1][k + 1][3];
            for (long[][][] table : dp) {
                for (long[][] row : table) {
                    for (long[] col : row) {
                        Arrays.fill(col, -1);
                    }
                }
            }

            // Take the minimum of both scenarios
            ans = Math.min(ans, rec(0, k, 0, 0, reversedArr));

            // Output result
            System.out.println(ans);
        }

        sc.close();
    }
}

**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
5 3
1 1000000000 1000000000 1000000000 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
#include <iostream>
#include <queue>
#include<map>
#include<algorithm>
#include<string>
#include <stack>
//#define int long long
typedef long long LL;
using namespace std;
typedef pair<LL, LL > PII;
typedef pair<long long , PII> que;//个数，价值
typedef pair<long long, PII> que;//价值，个数
stack<string>st;
/*LL a[200005];
LL b[200005];
map<long long, PII>mp;
priority_queue<que, vector<que>, greater<que>>qu;//时间，伤害，序号*/

map<LL, LL>ans;//价值，个数
LL se[100005], qe[100005];

using namespace std;
using ll = long long;
LL dep[500005];//i的深度
vector<LL>e[500005];//边
LL maxdep[500005];//每一个点所在树的最深长度
LL maxd=0;
LL deplive[500005];//每一层活了多少
LL maxans = 0;
//二叉树深度模板
void dfs(LL n, LL f)
{
	maxdep[n]=dep[n] = dep[f] + 1;
	maxd = max(maxd, dep[n]);
	deplive[dep[n]]++;
	for (auto i : e[n])
	{
		if (i == f)continue;
		dfs(i,n);
		maxdep[n] = max(maxdep[i],maxdep[n]);
	}

	if(maxdep[n]< dep[n]+1)
	deplive[dep[n] + 1]--;
}

int main() {
	LL n;
	cin >> n;//n条边

	while (n--)
	{

		LL t; cin >> t;
		for (LL i = 0; i < t-1; i++)
		{
			LL x, y; cin >> x >> y;
			e[x].push_back(y);
			e[y].push_back(x);
		}
		dep[0] = 0;
		maxdep[1] = 1;
		dfs(1, 0);
		LL live = 0;
		for (LL i = 1; i <= maxd; i++)
		{
			live += deplive[i];
			maxans = max(maxans, live);
		}
		cout <<t- maxans<<endl;

		for (LL i = 0; i <= t; i++)
		{
			dep[i] = 0;
			maxdep[i] = 0;
			deplive[i] = 0;
			e[i].clear();
		}maxd = 0;

	}

	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 create a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases or scenarios that might cause the code to fail. Here are the steps to think through the problem:

1. **Understand the Bug**: The buggy code calculates the depth of each node and the maximum depth of the subtree rooted at each node. It then tries to find the maximum number of nodes at the same depth and subtracts this from the total number of nodes to get the minimum number of operations. The issue might be in how it handles the depth calculation or the pruning logic.

2. **Edge Cases**: Consider edge cases such as:
   - Trees with very few nodes.
   - Trees with a very deep structure.
   - Trees with a very wide structure.
   - Trees where the root has multiple children with different depths.
   - Trees where the leaves are not at the same depth initially but can be pruned to be at the same depth.

3. **Small-Scale Test Input**: Since the test input should be small-scale, we can create a tree with a few nodes that might expose the bug. For example, a tree where the root has multiple children with different depths.

4. **Example Test Input**: Let's create a tree with 7 nodes where the root has three children, and each child has a different depth.

Here is the test input:

```plaintext
1
7
1 2
1 3
1 4
2 5
3 6
4 7
```

Explanation:
- The tree has 7 nodes.
- Node 1 is the root.
- Node 1 has three children: Node 2, Node 3, and Node 4.
- Node 2 has one child: Node 5.
- Node 3 has one child: Node 6.
- Node 4 has one child: Node 7.
- The depths of the leaves are:
  - Node 5: Depth 2
  - Node 6: Depth 2
  - Node 7: Depth 2

This tree should require 2 operations to make all leaves at the same distance from the root. The expected output is 2. This test input is small and valid, and it should help expose any issues in the depth calculation or pruning logic 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: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
vector<int>v[500001];
vector<bool>vis(500001);
vector<long long int> cur(500001);
vector<long long int> ma(500001);
int dfs(int node, int len){
    if(vis[node]){
        return 0;
    }
    int maxe = len;
    vis[node] = true;
    if(node!=0){
        cur[len]++;
    }
    for(auto x: v[node]){
        maxe = max(maxe, dfs(x, len+1));
    }
    if(node!=0){
        ma[maxe]++;
    }
    return maxe;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    std::cin >> t;
    while(t--){
        int n;
        std::cin >> n;
        int x,y;
        for(int i=0;i<n;i++){
            v[i].clear();
            vis[i] = false;
            cur[i]=0;
            ma[i]=0;
        }
        for(int i=0;i<n-1;i++){
            std::cin >> x >> y;
            x--;
            y--;
            v[x].push_back(y);
            v[y].push_back(x);
        }
        dfs(0,0);
        for(int i=1;i<=n;i++){
            ma[i] = ma[i] + ma[i-1];
        }
        for(int i=n-1;i>=0;i--){
            cur[i] = cur[i] + cur[i+1];
        }
        long long int ans = n-1;
        for(int i=1;i<=n;i++){
            ans = min(ans, ma[i-1]+cur[i+1]);
        }
        std::cout << ans << '\n';
    }
    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
3
1 2
1 3
4
1 2
1 3
1 4
5
1 2
1 3
1 4
1 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
#pragma GCC optimize("03")
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

/*```````````````````````````****Defined macros*****``````````````````````````*/
const int mod = 1e9 + 7;
#define ll long long
#define endl "\n"
#define ull unsigned long long
#define pb push_back
#define yes cout << "YES" << endl;
#define no cout << "NO" << endl;
#define set_bits __builtin_popcountll
#define all(x) x.begin(), x.end()
#define vll vector<ll>
#define vii vector<int>
#define tll takein<ll>
#define tii takein<int>
#define loop(i, a, b) for (int i = a; i < b; i++)

template <class T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <class K, class V>
using ordered_map = tree<K, V, less<K>, rb_tree_tag, tree_order_statistics_node_update>;

/*```````````````````````````````//DEBUGGER``````````````````````````````````*/
#ifndef ONLINE_JUDGE
#define _debug(x)      \
    cerr << #x << " "; \
    _print(x);         \
    cerr << endl;
#else
#define _debug(x)
#endif
template <typename T>
void _print(T x) { cerr << x; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p)
{
    cerr << "{";
    _print(p.first);
    cerr << ",";
    _print(p.second);
    cerr << "}";
}
template <class T>
void _print(vector<T> v)
{
    cerr << "[ ";
    for (T i : v)
    {
        _print(i);
        cerr << " ";
    }
    cerr << "]";
}
template <class T>
void _print(set<T> v)
{
    cerr << "[ ";
    for (T i : v)
    {
        _print(i);
        cerr << " ";
    }
    cerr << "]";
}
template <class T>
void _print(multiset<T> v)
{
    cerr << "[ ";
    for (T i : v)
    {
        _print(i);
        cerr << " ";
    }
    cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v)
{
    cerr << "[ ";
    for (auto i : v)
    {
        _print(i);
        cerr << " ";
    }
    cerr << "]";
}
template <typename T>
vector<T> takein(int n)
{
    vector<T> A;
    for (int i = 0; i < n; i++)
    {
        T x;
        cin >> x;
        A.push_back(x);
    }
    return A;
}
template <typename T>
vector<T> presum(vector<T> &A)
{
    vector<T> ans(A.size() + 1, 0);
    for (int i = 1; i <= A.size(); i++)
    {
        ans[i] = ans[i - 1] + A[i - 1];
    }
    return ans;
}
template <typename T>
void idxocc(vector<T> A, vector<vector<T>> idx)
{
    for (int i = 0; i < A.size(); i++)
    {
        T a;
        cin >> a;
        idx[a].push_back(i);
    }
}
template <typename T>
void print_arr(vector<T> ans)
{
    for (auto i : ans)
    {
        cout << i << " ";
    }
    cout << endl;
}
ll binExp(int N, int R)
{
    ll ans = 1;
    while (R > 0)
    {
        if (R % 2 == 1)
        {
            ans = (ans * 1LL * N);
        }
        N = (N * 1LL * N);
        R = R >> 1;
    }
    return ans;
}
ll modInv(int N, int R, int Mod)
{
    ll ans = 1;
    while (R > 0)
    {
        if (R % 2 == 1)
        {
            ans = (ans * 1LL * N) % Mod;
        }
        N = (N * 1LL * N) % Mod;
        R = R >> 1;
    }
    return ans % Mod;
}

/*```````````````````````````````//predefined f(x)``````````````````````````````````*/

int x_Nx = 1e5 + 1;
vector<bool> prime(x_Nx, 1);
void primality()
{
    prime[1] = prime[0] = 0;
    for (int i = 2; i * i <= x_Nx; i++)
    {
        if (prime[i] == 1)
        {
            for (int j = i * i; j <= x_Nx; j = j + i)
            {
                prime[j] = 0;
            }
        }
    }
    return;
}
// use binsea snippet for Binary_search code

// ceil(A/B)=(A+B-1)/B
// log2(A)==Highest set bit in A
/*```````````````````````````````//solution``````````````````````````````````*/
unordered_map<int, list<int>> adj;
vector<int> parent;
vector<int> depth, subtree, vis;

void parentDfs(int vertex)
{
    vis[vertex] = 1;
    for (auto &i : adj[vertex])
    {
        if (!vis[i])
        {
            parent[i] = vertex;
            parentDfs(i);
        }
    }
    return;
}

int depthdfs(int vertex)
{
    vis[vertex]=1;
    for (auto &i : adj[vertex])
    {
        if (!vis[i])
        {
            depth[vertex] = max(depthdfs(i), depth[vertex]);
        }
    }
    return depth[vertex] + 1;
}

int subtreeDfs(int vertex)
{
    int c = 0;
    vis[vertex]=1;

    for (auto &i : adj[vertex])
    {
        if (!vis[i])
        {
            c += subtreeDfs(i);
        }
    }
    subtree[vertex] = c;
    return c + 1;
}
bool isLeaf(int &vertex)
{
    if (depth[vertex]==0)
    {
        return true;
    }
    return false;
}
int superparent(int vertex)
{
    while (parent[vertex] != 1)
    {
        vertex = parent[vertex];
    }
    return vertex;
}

void solve()
{
    // solve the question here
    int n;
    cin >> n;
    adj.clear();
    // intitialisations
    parent.clear();
    depth.clear();
    subtree.clear();
    vis.clear();
    parent.resize(n + 1, 0);
    depth.resize(n + 1, 0);
    subtree.resize(n + 1, 0);
    parent[1] = -1;
    vis.resize(n + 1, 0);

    for (int i = 0; i < n - 1; i++)
    {
        int x, y;
        cin >> x >> y;
        adj[x].pb(y);
        adj[y].pb(x);
    }

    parentDfs(1);
    vis.clear();
    vis.resize(n + 1, 0);
    depthdfs(1);
    vis.clear();
    vis.resize(n + 1, 0);
    subtreeDfs(1);
    _debug(depth);
    _debug(subtree);
    _debug(parent);
    vis.clear();
    vis.resize(n + 1, 0);
    vis[1]=1;
    queue<int> q;
    q.push(1);
    int until = 0;
    int ans = subtree[1];
    while (!q.empty())
    {
        int x = q.size();
        int sz = 0;
        int temp = 0;
        set<int> nalle;
        while (x--)
        {
            int f = q.front();
            q.pop();
            for (auto &i : adj[f])
            {
                if (!vis[i])
                {
                    _debug(i);
                    sz += subtree[i];
                    if (isLeaf(i))
                    {

                        if (depth[parent[i]] == 1)
                        {
                            nalle.insert(parent[i]);
                        }
                        else
                        {
                            int z = superparent(i);
                            subtree[z]--;
                            temp++;
                        }
                    }
                    q.push(i);
                }
                vis[i]=1;
            }
        }

        ans = min(ans, sz + until);
        _debug(ans);
        while (!nalle.empty())
        {
            int x = *nalle.begin();
            _debug(x);
            nalle.erase(x);
            int y = superparent(x);
            until += subtree[y] + 1;
        }
        until += temp;
        _debug(until);
    }
    cout << ans << endl;
}

int main()
{
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    srand(chrono::high_resolution_clock::now().time_since_epoch().count());
    /*#ifndef ONLINE_JUDGE
    freopen("Error.txt", "w", stderr);
    #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 create a test input that exposes the bug in the provided code, we need to focus on edge cases or specific scenarios that are likely to cause the algorithm to fail. Here are some considerations:

1. **Single Path Tree**: A tree that is a single path from the root to a leaf.
2. **Balanced Tree**: A tree where all leaves are at the same depth initially.
3. **Unbalanced Tree**: A tree where leaves are at different depths, but the optimal solution involves removing specific edges to make all leaves at the same depth.
4. **Tree with Multiple Levels of Leaves**: A tree with multiple levels of leaves, where some levels have only a few leaves.

Given these considerations, let's create a test input for an unbalanced tree where the optimal solution involves removing specific edges to make all leaves at the same depth.

### Test Input
```plaintext
1
7
1 2
1 3
2 4
2 5
4 6
4 7
```

### Explanation
- **Number of Test Cases**: 1
- **Number of Nodes**: 7
- **Edges**:
  - (1, 2)
  - (1, 3)
  - (2, 4)
  - (2, 5)
  - (4, 6)
  - (4, 7)

### Expected Output
The expected output for this test case is `2`.

### Reasoning
- **Initial Tree Structure**:
  - Root (1) has two children: 2 and 3.
  - Node 2 has two children: 4 and 5.
  - Node 4 has two children: 6 and 7.
  - Node 3 is a leaf at depth 1.
  - Nodes 5, 6, and 7 are leaves at depth 3.

- **Optimal Solution**:
  - To make all leaves at the same depth, we need to remove nodes 3 and 5.
  - After removing nodes 3 and 5, the remaining leaves (6 and 7) are at depth 3 from the root.

This test input should expose any issues in the algorithm's handling of unbalanced trees and the correct identification of nodes to remove to achieve the goal.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
class Node:
    def __init__(self,val,h=0):
        self.parent=None
        self.val=val
        self.height=h
        self.nchild=0

def update(tree,store_h,ind):
    if(len(tree[ind])==1):
        return store_h
    h=store_h[ind].height
    store_h[ind].nchild=(len(tree[ind])-1)
    for i in tree[ind]:
        if(i==store_h[ind].parent):
            continue
        store_h[i].height=h+1
        store_h[i].parent=ind
        store_h=update(tree,store_h,i)
        store_h[ind].nchild+=store_h[i].nchild

    return store_h

if __name__ == '__main__':
    t=int(input())
    while(t>0):
        t-=1
        n=int(input())
        tree=[[] for i in range(n+1)]
        for i in range(n-1):
            c,d=map(int,input().split())
            tree[c].append(d)
            tree[d].append(c)

        store_h=[Node(i) for i in range(0,n+1)]
        for i in tree[1]:
            store_h[i].parent=1
        store_h=update(tree,store_h,1)
        store_h[1].nchild=len(tree[1])
        for i in tree[1]:
            store_h[1].nchild+=store_h[i].nchild
        heights={}
        heights[0]=[1,len(tree[1])]
        for i in range(2,n+1):
            h1=store_h[i].height
            if h1 not in heights:
                heights[h1]=[1,store_h[i].nchild]
            else:
                heights[h1][0]+=1
                heights[h1][1]+=store_h[i].nchild
            # print(i,store_h[i].height,store_h[i].parent,store_h[i].nchild)
        m1=[0,5*10**5]
        fp=-1
        for (key,val) in heights.items():
            # print(key)
            if(val[0]>m1[0]):
                if(val[1]<m1[1]):
                    fp=key
                    m1=val
        nc=0
        for i in range(1,n+1):
            if(store_h[i].height<fp and store_h[i].nchild==0):
                nc+=1
        ans=m1[1]+nc
        print(ans)
        # for i in range(1,n+1):
        #     if(len(tree[i])==1):

**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 code, we need to consider scenarios that might cause the algorithm to fail. The main issues in the code could be related to:

1. **Incorrect height calculation**: The code might not correctly update the heights of the nodes.
2. **Incorrect child count**: The code might not correctly count the number of children for each node.
3. **Incorrect pruning logic**: The code might not correctly determine the minimum number of operations needed to prune the tree.

Let's consider a small-scale test case that can expose these issues:

### Test Case 1: A Simple Tree with Uneven Depths
This test case will check if the code correctly handles a tree with nodes at different depths and ensures that all leaves are at the same distance from the root.

```plaintext
1
7
1 2
1 3
2 4
2 5
4 6
4 7
```

### Explanation:
- **Tree Structure**:
  - Node 1 is the root.
  - Node 2 and Node 3 are direct children of Node 1.
  - Node 4 and Node 5 are children of Node 2.
  - Node 6 and Node 7 are children of Node 4.
- **Expected Output**:
  - The tree has leaves at different depths: Node 3 at depth 1, Node 5 at depth 2, and Nodes 6 and 7 at depth 3.
  - To make all leaves at the same depth, we need to remove Node 3 and Node 5.
  - The minimum number of operations needed is 2.

### Test Case 2: A Linear Tree
This test case will check if the code correctly handles a linear tree where all nodes are in a single path.

```plaintext
1
6
1 2
2 3
3 4
4 5
5 6
```

### Explanation:
- **Tree Structure**:
  - Node 1 is the root.
  - Each node has exactly one child, forming a linear path.
- **Expected Output**:
  - The tree has leaves at different depths: Node 6 is at depth 5.
  - To make all leaves at the same depth, we need to remove all nodes except the last one.
  - The minimum number of operations needed is 5.

### Test Case 3: A Tree with Multiple Branches
This test case will check if the code correctly handles a tree with multiple branches and leaves at different depths.

```plaintext
1
9
1 2
1 3
2 4
2 5
3 6
3 7
4 8
4 9
```

### Explanation:
- **Tree Structure**:
  - Node 1 is the root.
  - Node 2 and Node 3 are direct children of Node 1.
  - Node 4 and Node 5 are children of Node 2.
  - Node 6 and Node 7 are children of Node 3.
  - Node 8 and Node 9 are children of Node 4.
- **Expected Output**:
  - The tree has leaves at different depths: Node 5 at depth 2, Node 6 and Node 7 at depth 2, and Node 8 and Node 9 at depth 3.
  - To make all leaves at the same depth, we need to remove Node 5.
  - The minimum number of operations needed is 1.

### Final Test Input
Combining the above test cases, we get:

```plaintext
3
7
1 2
1 3
2 4
2 5
4 6
4 7
6
1 2
2 3
3 4
4 5
5 6
9
1 2
1 3
2 4
2 5
3 6
3 7
4 8
4 9
```

This test input 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: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
from collections import deque, defaultdict
import sys
input = sys.stdin.read

def solve():
    data = input().split()

    index = 0
    t = int(data[index])
    index += 1
    results = []

    for _ in range(t):
        n = int(data[index])
        index += 1

        # Adjacency list representation of the tree
        adj = defaultdict(list)

        for __ in range(n - 1):
            u = int(data[index])
            v = int(data[index + 1])
            index += 2
            adj[u].append(v)
            adj[v].append(u)

        # Step 1: Calculate the depth of each node using BFS
        depth = [-1] * (n + 1)
        max_depth = 0

        queue = deque([1])
        depth[1] = 0

        while queue:
            node = queue.popleft()
            current_depth = depth[node]

            for neighbor in adj[node]:
                if depth[neighbor] == -1:
                    depth[neighbor] = current_depth + 1
                    max_depth = max(max_depth, depth[neighbor])
                    queue.append(neighbor)

        # Step 2: Identify leaves and their depths
        leaves_at_depth = [0] * (max_depth + 1)
        total_leaves = 0

        for i in range(2, n + 1):
            if len(adj[i]) == 1:  # Node i is a leaf
                leaves_at_depth[depth[i]] += 1
                total_leaves += 1

        # Step 3: Find the minimum number of removals required
        min_removals = total_leaves - leaves_at_depth[max_depth]
        results.append(min_removals)

    print('\n'.join(map(str, results)))

# Call the solve function to execute the solution
solve()

**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 code, we need to craft a test case that highlights the issue in the algorithm. The bug in the code is that it only considers the maximum depth and the number of leaves at that depth, but it does not correctly handle the scenario where there are multiple depths with leaves, and the optimal solution might involve removing leaves from different depths.

### Step-by-Step Reasoning:

1. **Understanding the Bug:**
   - The code calculates the minimum number of removals as `total_leaves - leaves_at_depth[max_depth]`.
   - This approach assumes that all leaves at the maximum depth should be kept, and all other leaves should be removed. However, this is not always the optimal solution.

2. **Creating a Test Case:**
   - We need a tree where the optimal solution involves removing leaves from different depths, not just from the maximum depth.
   - A good example is a tree where leaves are at multiple depths, and the optimal solution involves removing leaves from a depth that is not the maximum depth.

3. **Example Test Case:**
   - Consider a tree with 7 nodes where leaves are at depths 2 and 3.
   - The optimal solution would involve removing leaves from depth 2 to ensure all remaining leaves are at depth 3.

### Test Case:
```plaintext
1
7
1 2
1 3
2 4
2 5
3 6
3 7
```

### Explanation:
- **Tree Structure:**
  - Node 1 is the root.
  - Node 1 has children 2 and 3.
  - Node 2 has children 4 and 5.
  - Node 3 has children 6 and 7.
- ** Depths:**
  - Depth 1: Node 1 (root)
  - Depth 2: Nodes 2 and 3
  - Depth 3: Nodes 4, 5, 6, and 7
- **Leaves:**
  - Nodes 4, 5, 6, and 7 are leaves at depth 3.
  - The optimal solution is to remove the leaves at depth 2 (nodes 2 and 3) to ensure all remaining leaves are at depth 3.

### Expected Output:
- The minimum number of operations needed is 2.

This test case should expose the bug in the code, as the code will incorrectly assume that all leaves at the maximum depth should be kept, leading to an incorrect result.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
from collections import defaultdict, deque
import sys

def min_ops_to_equal_leaves(t_cases):
    res = []

    for n, edges in t_cases:
        tree = defaultdict(list)
        for u, v in edges:
            tree[u].append(v)
            tree[v].append(u)

        depth_count = defaultdict(int)
        vis = [False] * (n + 1)
        q = deque([(1, 0)])
        vis[1] = True

        while q:
            node, depth = q.popleft()
            is_leaf = True

            for nbr in tree[node]:
                if not vis[nbr]:
                    vis[nbr] = True
                    q.append((nbr, depth + 1))
                    is_leaf = False

            if is_leaf:
                depth_count[depth] += 1

        if depth_count:
            max_depth = max(depth_count.keys())
            total_leaves = sum(depth_count.values())
            leaves_at_max_depth = depth_count[max_depth]
            ops = total_leaves - leaves_at_max_depth
        else:
            ops = 0

        res.append(ops)

    return res

# Example usage
t = int(sys.stdin.readline())
t_cases = []
for _ in range(t):
    n = int(sys.stdin.readline())
    edges = [tuple(map(int, sys.stdin.readline().split())) for _ in range(n - 1)]
    t_cases.append((n, edges))

results = min_ops_to_equal_leaves(t_cases)
for r in results:
    print(r)

**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
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
import java.util.*;
public class treeprun
{
    Scanner sc = new Scanner(System.in);
    static final int inf = Integer.MAX_VALUE;
    static final int mod = 1000000007;
    static final int modl = 1000000007;
    static int maxd=0;
    static int count=0;
    static int max(int a, int b)
    {
        return a>b?a:b;
    }
    static int min(int a, int b)
    {
        return a<b?a:b;
    }
    static void in(int n, int arr[], Scanner sc)
    {
        for(int i=0;i<n;i++)
        arr[i]=sc.nextInt();
    }
    static int in(Scanner sc)
    {
        return sc.nextInt();
    }
    static void println(long a)
    {
        System.out.println(a);
    }
    static void println(double a)
    {
        System.out.println(a);
    }
    static void println(String a)
    {
        System.out.println(a);
    }
    static void print(long a)
    {
        System.out.print(a+" ");
    }
    static void print(double a)
    {
        System.out.print(a+" ");
    }
    static void print(String a)
    {
        System.out.print(a+" ");
    }
    static void println()
    {
        System.out.println();
    }
    static void printarr(int arr[])
    {
        for(int i=0;i<arr.length;i++)
        {
            System.out.print(arr[i]+" ");
        }
        System.out.println();
    }
    static void sort(int a[])
    {
        Arrays.sort(a);
    }
    static void revsort(int arr[])
    {
        Arrays.sort(arr);
        for (int i = 0; i < arr.length / 2; i++)
        {
           int temp = arr[i];
           arr[i] = arr[arr.length - 1 - i];
           arr[arr.length - 1 - i] = temp;
        }
    }
    static void puts(HashMap<Integer, Integer> map, int n)
    {
        map.put(n,map.getOrDefault(n,0)+1);
    }
    static void dfs1(int current, int parent, ArrayList<ArrayList<Integer>> list, HashMap<Integer, Integer> map, int cd)
    {
        if(cd>maxd)maxd=cd;
        map.put(cd,map.getOrDefault(cd,0)+1);
        for(int x:list.get(current))
        {
            if(x==parent)continue;
            dfs1(x,current,list,map,cd+1);
        }
    }

    static void dfs2(int current, int parent, ArrayList<ArrayList<Integer>> list, int val[], int cd)
    {
        count++;
        val[cd]+=count;
        for(int x:list.get(current))
        {
            if(x==parent)continue;

            dfs2(x,current,list,val,cd+1);
        }
        count=0;
    }
    static void solve(Scanner sc)
    {
        int n=in(sc);
        ArrayList<ArrayList<Integer>> list = new ArrayList<>();
        for(int i=0;i<=n;i++)
        {
            list.add(new ArrayList<Integer>());
        }
        for(int i=0;i<n-1;i++)
        {
            int u=in(sc);
            int v=in(sc);
            list.get(u).add(v);
            list.get(v).add(u);
        }

        HashMap<Integer, Integer> map = new HashMap<>();
        dfs1(1,0,list,map,1);

        int greater[] = new int[maxd+1];
        greater[maxd]=0;
        for(int i=maxd-1;i>=1;i--)
        {
            greater[i]=greater[i+1]+map.get(i+1);
        }
        int val[] = new int[n+1];

        dfs2(1,0,list,val,1);
        count=0;
        int summin=inf;
        for(int i=1;i<=maxd;i++)
        {
            int sum=greater[i]+n-greater[i]-val[i];
            if(sum<summin)summin=sum;
        }
        println(summin);
        maxd=0;
    }
    public static void main(String args[])
    {
        Scanner sc = new Scanner(System.in);
        int cases = sc.nextInt();
        while(cases-->0)solve(sc);
    }
}

**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 maximize the failure probability. Here are the steps to think through this:

1. **Understanding the Bug**: The provided code has a complex structure with multiple DFS traversals and calculations. The main issue might lie in how it handles the depth of nodes and the count of nodes at each depth.

2. **Edge Cases**:
   - **Small Tree**: A small tree with a few nodes to ensure the basic logic is correct.
   - **Linear Tree**: A tree where nodes form a straight line to test the depth handling.
   - **Balanced Tree**: A balanced tree to test the symmetry and depth distribution.
   - **Tree with Multiple Leaves at Different Depths**: A tree with leaves at different distances from the root to test the pruning logic.

3. **Small-Scale Test Input**:
   - We will create a tree with 7 nodes, which is small enough to be manageable but large enough to test the logic.

4. **Test Input**:
   - **Tree Structure**: Rooted at node 1, with leaves at different depths.
   - **Expected Output**: The minimum number of operations needed to make all leaves at the same distance from the root.

Here is the test input:

```plaintext
1
7
1 2
1 3
2 4
2 5
4 6
4 7
```

This test input represents a tree with the following structure:

```
    1
   / \
  2   3
 / \
4   5
 \
  6
 \
  7
```

In this tree:
- Node 3 is a leaf at depth 1.
- Nodes 5 and 7 are leaves at depth 3.
- Nodes 6 is a leaf at depth 4.

The expected output is `2`, as removing edges `(1, 3)` and `(4, 6)` will result in a tree where all leaves (nodes 5 and 7) are at the same distance from the root (node 1), which is 3.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Tree Pruning
time_limit_ms: 3000
memory_limit_mb: 256
Description: [t+pazolite, ginkiha, Hommarju - Paved Garden](https://soundcloud.com/fractalex-gd/ginkiha-paved-garden-little)

You are given a tree with $n$ nodes, rooted at node $1$. In this problem, a leaf is a non-root node with degree $1$.

In one operation, you can remove a leaf and the edge adjacent to it (possibly, new leaves appear). What is the minimum number of operations that you have to perform to get a tree, also rooted at node $1$, where all the leaves are at the same distance from the root?

Input: Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 10^4$). The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($3 \leq n \leq 5 \cdot 10^5$) — the number of nodes.

Each of the next $n-1$ lines contains two integers $u$, $v$ ($1 \leq u, v \leq n$, $u \neq v$), describing an edge that connects $u$ and $v$. It is guaranteed that the given edges form a tree.

It is guaranteed that the sum of $n$ over all test cases does not exceed $5 \cdot 10^5$.

Output: For each test case, output a single integer: the minimum number of operations needed to achieve your goal.

Examples:
input:
3
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
12 9
1 6
6 14
9 11
8 7
3 5
13 5
6 10
13 15
13 6
14 12
7 2
8 1
1 4
output:
2
2
5

Note: In the first two examples, the tree is as follows:

![](CDN_BASE_URL/754511fbdb88995aaa49733a1fe75dfc)

In the first example, by removing edges $(1, 3)$ and $(2, 5)$, the resulting tree has all leaves (nodes $6$ and $7$) at the same distance from the root (node $1$), which is $3$. The answer is $2$, as it is the minimum number of edges that need to be removed to achieve the goal.

In the second example, removing edges $(1, 4)$ and $(5, 7)$ results in a tree where all leaves (nodes $4$ and $5$) are at the same distance from the root (node $1$), which is $2$.

Buggy Code:
import java.util.*;
public class TreePruning { static ArrayList<ArrayList<Integer>>list=new ArrayList<>();static int dp[][];
static boolean visit[];
    public static void main(String args[])
    {
        Scanner in=new Scanner(System.in);
        int t=in.nextInt();
        while(t-->0)
        {
            int n=in.nextInt();
           list=new ArrayList<>();
           for(int i=0;i<=n;i++)
           list.add(new ArrayList<>());
           dp=new int[n+1][2];visit=new boolean[n+1];
           for(int i=0;i<n-1;i++)
           {
            int x=in.nextInt(),y=in.nextInt();
            list.get(x).add(y);list.get(y).add(x);
           }
           dfs(1,1);
           Arrays.sort(dp,(a,b)->a[0]-b[0]);int pre[]=new int[n+1];
           for(int i=1;i<n+1;i++)
           {
                if(dp[i][1]==1)pre[i]=pre[i-1]+1;
                else pre[i]=pre[i-1];
                // System.out.println(pre[i]+"o");
           }long ans=100000000;
           long s[]=new long[n+2];
           s[n+1]=0;
           s[n]=dp[n][0];
           for(int i=n-1;i>=1;i--){
           s[i]=dp[i][0]+s[i+1];}
           for(int i=1;i<=n;i++)
           {
            int left=1,right=n,index=n+1;int x=dp[i][0];
            while(left<=right)//just greater
            {
                int mid=(left+right)/2;
                if(dp[mid][0]>x)
                {index=mid;right=mid-1;}
                else left=mid+1;
            }left=1;right=n;int index2=1;
            while(left<=right)//just smaller
            {
                int mid=(left+right)/2;
                if(dp[mid][0]>=x)
                {right=mid-1;}
                else {index2=mid;left=mid+1;}
            }
            // System.out.println(index+" "+index2+" "+s[index]+" "+pre[index2]);
            ans=Math.min(ans,-(long)x*(long)(n+1-index)+s[index]+pre[index2]);
            // System.out.println(ans);
           }System.out.println(ans);
        }
    }
    public static void dfs(int parent,int c)
    {
        if(visit[parent])return;
        dp[parent][0]=c;visit[parent]=true;
        if(list.get(parent).size()==1){dp[parent][1]=1;}
           for(int i:list.get(parent))
           {
            dfs(i,c+1);
           }
           return;
}}

**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
7
1 2
1 3
2 4
2 5
4 6
4 7
7
1 2
1 3
1 4
2 5
3 6
5 7
15
1 2
1 3
1 4
2 5
3 6
3 7
4 8
4 9
5 10
5 11
6 12
6 13
7 14
7 15
```