**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (hard version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $$$r \geq l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r \geq l+k-1$$$) — the bounds of the query.

It is guaranteed the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

Examples:
input:
3
7 5 3
1 2 3 2 1 2 3
1 7
2 7
3 7
8 4 2
4 3 1 1 2 4 3 2
3 6
1 5
5 4 2
4 5 1 2 3
1 4
1 5
output:
6
5
2
2
5
2
3

Note: In the second query of the first testcase, we calculate the following function values:

- $$$f([2,3,2,1,2])=3$$$ because Yunli can set $$$b_3=4$$$, $$$b_4=5$$$, and $$$b_5=6$$$, making a consecutive subarray of size $$$5$$$ in $$$3$$$ moves.
- $$$f([2,3,2,1,2,3])=2$$$ because we can set $$$b_3=0$$$ and $$$b_2=-1$$$, making a consecutive subarray of size $$$5$$$ in $$$2$$$ moves (starting at position $$$2$$$)

The answer to this query is $$$3+2=5$$$.

Buggy Code:
t = int(input())
for _ in range(t):
    n, k, q = map(int, input().split())
    a = list(map(int, input().split()))
    max_len = [0]*n  # max_len[i] stores the maximum length ending at position i
    current_len = 1
    max_len[0] = 1
    for i in range(1, n):
        if a[i] == a[i-1] +1:
            current_len +=1
        else:
            current_len =1
        max_len[i] = max(max_len[i-1], current_len)
    # Build prefix sums of max_len
    prefix_max_len = [0]*n
    prefix_max_len[0] = max_len[0]
    for i in range(1, n):
        prefix_max_len[i] = max_len[i]
    # Process queries
    for _ in range(q):
        l, r = map(int, input().split())
        l -=1  # 0-based indexing
        total = 0
        for j in range(l + k -1, r):
            # Find the maximum length of increasing sequence in [l, j]
            m = 1
            current_len =1
            for i in range(l+1, j+1):
                if a[i] == a[i-1] +1:
                    current_len +=1
                else:
                    current_len =1
                m = max(m, current_len)
            total += max(0, k - m)
        print(total)

**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 2
1 2 3 1 2
1 5
2 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (hard version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $$$r \geq l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r \geq l+k-1$$$) — the bounds of the query.

It is guaranteed the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

Examples:
input:
3
7 5 3
1 2 3 2 1 2 3
1 7
2 7
3 7
8 4 2
4 3 1 1 2 4 3 2
3 6
1 5
5 4 2
4 5 1 2 3
1 4
1 5
output:
6
5
2
2
5
2
3

Note: In the second query of the first testcase, we calculate the following function values:

- $$$f([2,3,2,1,2])=3$$$ because Yunli can set $$$b_3=4$$$, $$$b_4=5$$$, and $$$b_5=6$$$, making a consecutive subarray of size $$$5$$$ in $$$3$$$ moves.
- $$$f([2,3,2,1,2,3])=2$$$ because we can set $$$b_3=0$$$ and $$$b_2=-1$$$, making a consecutive subarray of size $$$5$$$ in $$$2$$$ moves (starting at position $$$2$$$)

The answer to this query is $$$3+2=5$$$.

Buggy Code:
from collections import Counter
import heapq
class SegmentTree:
    def __init__(self, n):
        self.n = n
        self.data = [-1] * (4 * n)
        self.lazy = [0] * (4 * n)

    def update(self, l, r, val, node=1, start=0, end=None):
        if end is None:
            end = self.n - 1

        if self.lazy[node] != 0:
            self.data[node] = self.lazy[node] * (end - start + 1)
            if start != end:
                self.lazy[2 * node] = self.lazy[node]
                self.lazy[2 * node + 1] = self.lazy[node]
            self.lazy[node] = 0

        if start > r or end < l:
            return

        if start >= l and end <= r:
            self.data[node] = val * (end - start + 1)
            if start != end:
                self.lazy[2 * node] = val
                self.lazy[2 * node + 1] = val
            return

        mid = (start + end) // 2
        self.update(l, r, val, 2 * node, start, mid)
        self.update(l, r, val, 2 * node + 1, mid + 1, end)
        self.data[node] = self.data[2 * node] + self.data[2 * node + 1]

    def query(self, l, r, node=1, start=0, end=None):
        if end is None:
            end = self.n - 1

        if self.lazy[node] != 0:
            self.data[node] = self.lazy[node] * (end - start + 1)
            if start != end:
                self.lazy[2 * node] = self.lazy[node]
                self.lazy[2 * node + 1] = self.lazy[node]
            self.lazy[node] = 0

        if start > r or end < l:
            return 0

        if start >= l and end <= r:
            return self.data[node]

        mid = (start + end) // 2
        left_query = self.query(l, r, 2 * node, start, mid)
        right_query = self.query(l, r, 2 * node + 1, mid + 1, end)
        return left_query + right_query

for _ in range(int(input())):
    n, k, q = list(map(int, input().split()))
    nums = list(map(int, input().split()))
    c = Counter()
    if k == 1:
        for i in range(q):
            print(0)
    heap = []
    dp = [0]*n
    # tree = SegmentTree(n)
    for i, num in enumerate(nums):
        c[num - i] += 1
        heapq.heappush(heap, (-c[num - i], num - i))
        if i >= k-1:
            while  -c[heap[0][1]] != heap[0][0]:
                heapq.heappop(heap)
            dp[i] = k  + heap[0][0]
            # tree.update(i, k  + heap[0][0])
            c[nums[i-k+1] -(i-k+1)] -= 1
    # print(dp)

    querie = [[] for i in range(n)]
    tree = SegmentTree(n)
    ans = [0]*q
    for i in range(q):
        l, r = list(map(int, input().split()))
        querie[l+k-2].append((r-1, i))
    stack = []
    next_small = [n-1]*n
    for i in range(n):
        while stack and dp[stack[-1]] > dp[i]:
            next_small[stack.pop()] = i-1
        stack.append(i)
    # print(dp, next_small)

    for i in range(n-1, -1,-1):
        tree.update(i, next_small[i], dp[i])
        for j, idx in querie[i]:
            ans[idx] = tree.query(i, j)
    for i in range(q):
        print(ans[i])

**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
10 5 3
1 2 3 4 5 6 7 8 9 10
1 10
2 10
3 10
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (hard version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $$$r \geq l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r \geq l+k-1$$$) — the bounds of the query.

It is guaranteed the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

Examples:
input:
3
7 5 3
1 2 3 2 1 2 3
1 7
2 7
3 7
8 4 2
4 3 1 1 2 4 3 2
3 6
1 5
5 4 2
4 5 1 2 3
1 4
1 5
output:
6
5
2
2
5
2
3

Note: In the second query of the first testcase, we calculate the following function values:

- $$$f([2,3,2,1,2])=3$$$ because Yunli can set $$$b_3=4$$$, $$$b_4=5$$$, and $$$b_5=6$$$, making a consecutive subarray of size $$$5$$$ in $$$3$$$ moves.
- $$$f([2,3,2,1,2,3])=2$$$ because we can set $$$b_3=0$$$ and $$$b_2=-1$$$, making a consecutive subarray of size $$$5$$$ in $$$2$$$ moves (starting at position $$$2$$$)

The answer to this query is $$$3+2=5$$$.

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

/**
 * @author Sri Vangaru
 */
public class YunlisSubarrayQueries {
    static BufferedReader br;
    static PrintWriter pw;
    static int MOD = (int) (1e9 + 7); // 998244353;
    static int[] di = {0, 0, 1, -1}, dj = {1, -1, 0, 0};

    public static void main(String[] args) throws IOException {
        br = new BufferedReader(new InputStreamReader(System.in));
        pw = new PrintWriter(System.out);

        int t = Integer.parseInt(br.readLine()); // 1;
        while (t-->0) {
            solve();
        }

        br.close(); pw.close();
    }

    static void solve() throws IOException {
        /*
         * we can start off by precomputing the answers for each size-k window.
         *      by the way, how can this be done? having a sequence of n elements a_1, a_2, ..., a_n
         *      be consecutively increasing is the same as having the sequence
         *      a_1 - 1, a_2 - 2, ..., a_i - i, ..., a_n - n consist of all the same elements. so, we
         *      can replace all of the elements with these, and then use a sliding window technique
         *      to compute the minimum changes needed: for a given window, the number of changes needed
         *      is the length of the window minus the largest frequency. this can be computed by having
         *      a multiset of the elements for the current window we are looking at and having a sorted
         *      multiset of the frequencies of the elements so that we can conveniently get the largest
         *      frequency. we need a multiset so we actually store all the information that we need and
         *      don't accidentally store only one instance of a frequency if there are multiple elements
         *      in the window with that frequency.
         * so, ans[r] will effectively equal f(a[r-k+1, r-k+2, ..., r]). now, for computing
         * the requested summation, we first off want to calculate the sum as i goes from l+k-1 to r of
         * f(a[l, l+1, l+2, ..., i]). consider an arbitrary array c. it's clear that, by a simple brute
         * force algorithm, f(c) is equal to the minimum of f(d) across all length-k subarrays called d
         * contained within c. therefore, what we want to calculate for each query is
         * the sum as i goes from l+k-1 to r of min(ans[l+k-1], ans[l+k], ..., ans[i]).
         * given that that is the problem that we wish to solve, let's take a closer look at the ans
         * array. this is generally applicable to any array (i.e., is not specific to the calculations
         * that resulted in the ans array), but if l was equal to 1 (or 0, if we are using 0-indexing),
         * then we are looking for the sum of all the prefix minimums up to the rth element.
         * now, the issue is that l is changing across all the queries. if we try to preprocess the
         * prefix minimum array, then the problem is that for a given range [li, ri] of a query, the
         * values of the prefix minimum array within that range could be influenced by some small values
         * to the left of that range, which should not impact the calculation for that specific query,
         * so we need some way of constructing the prefix minimum array intelligently where once we
         * compute the answer for a given query, it doesn't invalidate the calculations for any future
         * queries.
         * for this exact reason, we can process the queries in order of right-to-left values of l. for
         * each query, we compute the new prefix minimum array by making it so that for a given element
         * m_i at index i, all elements to the right of it that are greater than it until the first that
         * is less than it are set to m_i. this can be done using a lazy segment tree, and the sums can
         * also be taken using the same segment tree. for each element, the index that we need to update
         * until (which is some point to the right or on the element) can easily be found either using a
         * monotonic stack or binary search.
         */

        String[] str = br.readLine().split(" ");
        int n = Integer.parseInt(str[0]), k = Integer.parseInt(str[1]), q = Integer.parseInt(str[2]);
        int[] a = Arrays.stream(br.readLine().split(" ")).mapToInt(Integer::parseInt).toArray();
        int[] ans = new int[n];
        Multiset diffs = new Multiset(), freqs = new Multiset();
        for (int i = 0; i <= k-1; i++) {
            diffs.add(a[i]-i);
        }
        for (Map.Entry<Integer, Integer> entry : diffs.mset.entrySet()) {
            freqs.add(entry.getValue());
        }
        ans[k-1] = k - freqs.mset.lastKey();

        for (int r = k; r <= n-1; r++) {
            int adding = a[r]-r, removing = a[r-k]-r+k;
            freqs.remove(diffs.count(removing));
            diffs.remove(removing);
            if (diffs.count(removing) > 0) freqs.add(diffs.count(removing));

            if (diffs.count(adding) > 0) freqs.remove(diffs.count(adding));
            diffs.add(adding);
            freqs.add(diffs.count(adding));

            ans[r] = k - freqs.mset.lastKey();
        }
        // pw.println("ans = "+Arrays.toString(ans));

        Stack<Integer> monotonic = new Stack();
        int[] rangeRight = new int[n];
        for (int i = k-1; i <= n; i++) {
            int currVal = i==n ? 0 : ans[i];
            while (!monotonic.isEmpty() && ans[monotonic.peek()] > currVal) {
                int idx = monotonic.pop();
                rangeRight[idx] = i-1;
            }
            monotonic.add(i);
        }
        // pw.println("rangeright = "+Arrays.toString(rangeRight));

        long[][] queries = new long[q][4];
        for (int i = 0; i <= q-1; i++) {
            str = br.readLine().split(" ");
            int l = Integer.parseInt(str[0])-1, r = Integer.parseInt(str[1])-1;
            queries[i] = new long[] {i, l, r, 0};
        }
        Arrays.sort(queries, Comparator.comparingLong(o -> o[1])); // comparing by left bound of query

        int prevLpKm1 = n;
        Lazy st = new Lazy(new int[n]);
        for (int i = q-1; i >= 0; i--) {
            long l = queries[i][1], r = queries[i][2];
            for (int j = prevLpKm1-1; j >= l+k-1; j--) {
                st.update(j, rangeRight[j], 1, ans[j]);
            }
            prevLpKm1 = (int) l+k-1;
            queries[i][3] = st.sum((int)(l+k-1), (int)r);
        }
        Arrays.sort(queries, Comparator.comparingLong(o -> o[0])); // sorting queries in original order

        for (int i = 0; i < q; i++) {
            pw.println(queries[i][3]);
        }
    }

    // // // //

    static class Multiset { // ordered multiset
        TreeMap<Integer, Integer> mset = new TreeMap<>();

        public void add(int x) {
            if (mset.containsKey(x)) {
                mset.put(x, mset.get(x) + 1);
            } else {
                mset.put(x, 1);
            }
        }

        public void remove(int x) {
            mset.put(x, mset.get(x) - 1);
            if (mset.get(x) == 0) mset.remove(x);
        }

        public void add(int x, int q) {
            // q for quantity
            if (mset.containsKey(x)) {
                mset.put(x, mset.get(x) + q);
            } else {
                mset.put(x, q);
            }
        }

        public int count(int x) {
            if (!mset.containsKey(x)) {
                return 0;
            } else {
                return mset.get(x);
            }
        }
    }

    static class DSU {
        // disjoint-set/union-find data structure
        int[] parent;
        int[] size;

        public DSU(int length) {
            parent = new int[length];
            size = new int[length];
            for (int i = 0; i <= length - 1; i++) parent[i] = i;
            Arrays.fill(size, 1);
        }

        public int find(int x) {
            if (parent[x] == x) return x;
            else {
                parent[x] = find(parent[x]);
                return parent[x];
            }
        }

        public void union(int a, int b) {
            int c = find(a);
            int d = find(b);
            if (c != d) {
                int sum = size[c] + size[d];
                parent[d] = c;
                size[c] = sum;
                size[d] = sum;
            }
        }

        public int size(int x) {
            return size[x];
        }
    }

    static long gcd(long a, long b) {
        long B = Math.min(a, b);
        long A = Math.max(a, b);
        while (B != 0) {
            long temp = B;
            B = A % B;
            A = temp;
        }
        return A;
    }

    static ArrayList<Integer> primeFactorization(int n) {
        ArrayList<Integer> factors = new ArrayList<Integer>();
        for (int i = 2; i <= Math.sqrt(n); i++) {
            while (n % i == 0) {
                n /= i;
                factors.add(i);
            }
        }
        if (n != 1) {
            factors.add(n);
        }
        return factors;
    }

    static class SegmentTree {
        // this implementation is for doing range maximum queries by default
        // it's written recursively for now :|
        int n;
        long[] t;
        long[][] lz;

        public SegmentTree(int[] original) {
            n = original.length;
            t = new long[4*n];
            lz = new long[4*n][2]; // lz[i][0] == no update, lz[i][1] == set, lz[i][2] == add

            build(original, 1, 0, n-1);
        }

        public void build(int[] a, int vertex, int vl, int vr) {
            if (vl == vr) {
                t[vertex] = a[vl];
            } else {
                int vm = (vl + vr) / 2;
                build(a, 2*vertex, vl, vm);
                build(a, 2*vertex+1, vm+1, vr);
                t[vertex] = Math.max(t[2*vertex], t[2*vertex+1]);
            }
        }

        // l and r are the bounds of the query; vl and vr are the bounds of the current vertex
        public long query(int l, int r, int vertex, int vl, int vr) {
            if (l > r)
                return Long.MIN_VALUE / 2;
            if (vl == l && vr == r)
                return t[vertex];

            int vm = (vl + vr) / 2;
            return Math.max(
                query(l, Math.min(r, vm), 2*vertex, vl, vm),
                query(Math.max(l, vm+1), r, 2*vertex+1, vm+1, vr)
            );
        }

        public long query(int l, int r) {
            return query(l, r, 1, 0, n-1);
        }

        public void update(int ind, int newVal, int vertex, int vl, int vr) {
            if (vl == vr) {
                t[vertex] = newVal;
            } else {
                int vm = (vl + vr) / 2;
                if (ind <= vm)
                    update(ind, newVal, 2*vertex, vl, vm);
                else
                    update(ind, newVal, 2*vertex+1, vm+1, vr);
                t[vertex] = Math.max(t[2*vertex], t[2*vertex+1]);
            }
        }

        public void update(int ind, int newVal) {
            update(ind, newVal, 1, 0, n-1);
        }
    }

    static class Lazy {
        int n;
        long[] t;
        long[][] lz;

        public Lazy(int[] original) {
            n = original.length;
            t = new long[4*n];
            lz = new long[4*n][2]; // lz[i][0] == 0: no update, lz[i][0] == 1: set, lz[i][0] == 2: add

            build(original, 1, 0, n-1);
        }

        public void build(int[] a, int vertex, int vl, int vr) {
            if (vl == vr) {
                t[vertex] = a[vl];
            } else {
                int vm = (vl + vr) / 2;
                build(a, 2*vertex, vl, vm);
                build(a, 2*vertex+1, vm+1, vr);
                t[vertex] = t[2*vertex] + t[2*vertex+1];
            }
        }

        public void apply(int v, int vlen, long type, long val) {
            if (type == 2) { // add update
                if (lz[v][0] != 1) {
                    // no set update already there, so this just places an add update with the
                    // previous query's value plus the current query's value
                    lz[v][0] = 2;
                    lz[v][1] = lz[v][1] + val;
                } else {
                    // adds the current query's value to the previous set query's value
                    lz[v][0] = 1;
                    lz[v][1] = lz[v][1] + val;
                }
                t[v] += (long) vlen * val;
            } else if (type == 1) { // set update
                // type == 1
                // the set operation overrides any past updates
                lz[v][0] = 1;
                lz[v][1] = val;
                t[v] = (long) vlen * val;
            }
        }

        public void pushDown(int v, int vl, int vr) {
            int vm = (vl + vr) / 2;
            apply(2*v, vm-vl+1, lz[v][0], lz[v][1]);
            apply(2*v+1, vr-vm, lz[v][0], lz[v][1]);
            lz[v][0] = 0;
            lz[v][1] = 0;
        }

        // range update, but point updates just have l == r
        public void update(int l, int r, long type, long val, int v, int vl, int vr) {
            if (vr < l || vl > r) return;
            if (l <= vl && vr <= r) {
                apply(v, vr-vl+1, type, val);
                return;
            }

            pushDown(v, vl, vr);
            int vm = (vl + vr) / 2;
            update(l, r, type, val, 2*v, vl, vm);
            update(l, r, type, val, 2*v+1, vm+1, vr);
            t[v] = t[2*v] + t[2*v+1];
        }

        public void update(int l, int r, long type, long val) {
            update(l, r, type, val, 1, 0, n-1);
        }

        public long sum(int l, int r, int v, int vl, int vr) {
            if (vr < l || vl > r) return 0;
            if (l <= vl && vr <= r) return t[v];

            pushDown(v, vl, vr);
            int vm = (vl + vr) / 2;
            return sum(l, r, 2*v, vl, vm) + sum(l, r, 2*v+1, vm+1, vr);
        }

        public long sum(int l, int r) {
            return sum(l, r, 1, 0, n-1);
        }
    }

    static long sqrt(long n) {
        // calculates the floor of sqrt(n) in O(log(n))
        if (n == 0) return 0;
        else if (n < 0) throw new IllegalArgumentException("can't return imaginary numbers");
        long l = 1, r = n, sqrt = -1;
        while (l <= r) {
            long mid = (l + r) / 2;
            long factor = n / mid;
            if (mid <= factor) {
                sqrt = mid;
                l = mid + 1;
            } else {
                r = mid - 1;
            }
        }
        return sqrt;
    }

    static class Point {
        int i, j;

        public Point(int i, int j) {
            this.i = i;
            this.j = j;
        }

        // int x, y;

        // public Point(int x, int y) {
        //     this.x = x;
        //     this.y = y;
        // }
    }

    static class DSU2 {
        // 2-dimensional DSU
        // this may or may not be useless since a 2D array can just be mapped to a 1D array :/
        Point[][] parent;
        int[][] size;

        public DSU2(int r, int c) {
            parent = new Point[r][c];
            size = new int[r][c];
            for (int i = 0; i <= r-1; i++) {
                for (int j = 0; j <= c-1; j++) {
                    parent[i][j] = new Point(i, j);
                }
            }
            for (int i = 0; i <= r-1; i++) {
                Arrays.fill(size[i], 1);
            }
        }

        public Point find(int i, int j) {
            if (parent[i][j].i == i && parent[i][j].j==j) {
                return parent[i][j];
            } else {
                parent[i][j] = find(parent[i][j].i, parent[i][j].j);
                return parent[i][j];
            }
        }

        public void union(int ai, int aj, int bi, int bj) {
            Point c = find(ai, aj);
            Point d = find(bi, bj);
            if (c.i != d.i || c.j != d.j) {
                int sum = size[c.i][c.j] + size[d.i][d.j];
                parent[d.i][d.j] = c;
                size[c.i][c.j] = sum;
                size[d.i][d.j] = sum;
            }
        }

        public int size(int i, int j) {
            return size[i][j];
        }
    }

    static long taxiDist(long x1, long y1, long x2, long y2) {
        // Manhattan distance between two points in the Cartesian coordinate plane
        return Math.abs(x2 - x1) + Math.abs(y2 - y1);
    }

    static double dist(long x1, long y1, long x2, long y2) {
        // uses the Pythagorean theorem to calculate the distance between two points
        // in the Cartesian coordinate plane
        long a = x2 - x1;
        long b = y2 - y1;
        return Math.sqrt(Math.pow(a, 2) + Math.pow(b, 2));
        // return (long) (Math.pow(a, 2) + Math.pow(b, 2)); // to return squared distance
    }

    static <T> void generatePermutations(int k, T[] a) {
        if (k == 1) {
            System.out.println(Arrays.toString(a)); // change this to whatever is considered as outputting the permutation
        } else {
            generatePermutations(k-1, a);
            for (int i = 0; i <= k-2; i++) {
                if (k % 2 == 0) {
                    T temp = a[i];
                    a[i] = a[k-1];
                    a[k-1] = temp;
                } else {
                    T temp = a[0];
                    a[0] = a[k-1];
                    a[k-1] = temp;
                }
                generatePermutations(k-1, a);
            }
        }
    }

    static class Pair {
        int a, b;

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

        @Override
        public boolean equals(Object obj) {
            if (this == obj) return true;
            if (obj == null) return false;
            if (!(obj instanceof Pair)) return false;

            Pair other = (Pair) obj;
            return this.a == other.a && this.b == other.b;
        }

        @Override
        public int hashCode() {
            return (this.a+" "+this.b).hashCode();
        }
    }

    // this runs the Extended Euclidean Algorithm on inputs a and b,
    // and it returns a length-3 array in the form [g, x, y],
    // such that g = gcd(a, b) = ax + by
    static long[] bezouts(long a, long b) {
        if (b == 0) {
            return new long[] {a, 1, 0}; // in the last step of the EA, gcd = a, so x = 1 and y = 0
        }

        long q = a / b, r = a % b; // a = bq + r, where 0 <= r < b
        long[] result = bezouts(b, r); // the "x" and "y" here are from the next step of the EA
        long gcd = result[0], x = result[1], y = result[2]; // gcd = bx + ry
        return new long[] {gcd, y, x - y * q}; // r = a - bq --> gcd = ay + b(x - yq) --> (x, y) := (y, x - yq)
    }

    static class PrefixTree {
        class TrieNode {
            TrieNode[] children;
            int end;

            public TrieNode() {
                children = new TrieNode[26];
                end = 0;
            }
        }

        private TrieNode root;

        public PrefixTree() {
            root = new TrieNode();
        }

        public void insert(String str) {
            TrieNode curr = root;
            for (int i = 0; i <= str.length() - 1; i++) {
                int letter = str.charAt(i) - 'a';
                if (curr.children[letter] == null) {
                    curr.children[letter] = new TrieNode();
                }
                curr = curr.children[letter];
            }
            curr.end++;
        }

        public boolean search(String str) {
            TrieNode curr = root;
            for (int i = 0; i <= str.length() - 1; i++) {
                int letter = str.charAt(i) - 'a';
                if (curr.children[letter] == null) {
                    return false;
                }
                curr = curr.children[letter];
            }
            return curr.end > 0;
        }

        public boolean startsWith(String prefix) {
            TrieNode curr = root;
            for (int i = 0; i <= prefix.length() - 1; i++) {
                int letter = prefix.charAt(i) - 'a';
                if (curr.children[letter] == null) {
                    return false;
                }
                curr = curr.children[letter];
            }
            return true;
        }

        public void delete(String str) {
            _delete(str, this.root, -1);
        }

        public boolean _delete(String str, TrieNode node, int index) {
            // reached the last letter
            if (index + 1 == str.length()) {
                // if the last letter isn't marked with end > 0, then the
                // word doesn't exist in the trie
                if (node.end == 0) {
                    return false;
                }

                // unmarks the node as the end of a word
                node.end--;

                // if there are other copies of the same word in the trie,
                // then return false (meaning we shouldn't delete any nodes
                // higher on the path from the root)
                if (node.end > 0) {
                    return false;
                }

                // since this word is now no longer part of the trie, we check
                // if any other words use this word as a prefix; if so, then we
                // delete nothing higher on the path from the path, but if not,
                // then we indicate that we can set this node as null from the
                // perspective of the parent node
                for (int i = 0; i <= 25; i++) {
                    if (node.children[i] != null) {
                        return false;
                    }
                }
                return true;
            }

            // not on last letter
            int nextLetter = str.charAt(index + 1) - 'a';

            // if the next letter is not in the trie, then the word isn't in the trie
            if (node.children[nextLetter] == null) {
                return false;
            } else {
                // if we don't want to delete the next letter from the trie, we don't
                // want to delete anything higher on the path from the root to the
                // current node, and so this cascades up
                if (_delete(str, node.children[nextLetter], index + 1) == false) {
                    return false;
                }

                // delete the next letter from the trie
                node.children[nextLetter] = null;

                // now, we just check if the current node is needed for any other words
                // (i.e., if it's a prefix of any); if it is, then we don't want to delete
                // anything further, and if it isn't, then we tell the node above to delete
                // this current node
                for (int i = 0; i <= 25; i++) {
                    if (node.children[i] != null) {
                        return false;
                    }
                }
                return true;
            }
        }

        public void printTrie() {
            _printTrie(0, '.', this.root);
        }

        public void _printTrie(int tab, char c, TrieNode node) {
            for (int i = 1; i <= tab; i++) System.out.print(' ');
            System.out.println("(\"" + c + "\", " + node.end + ")");
            for (int i = 0; i <= 25; i++) {
                if (node.children[i] != null) {
                    _printTrie(tab + 2, (char) ('a' + i), node.children[i]);
                }
            }
        }
    }
}

**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
7 3 4
1 2 3 1 2 3 4
1 7
2 7
3 7
4 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (hard version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $$$r \geq l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r \geq l+k-1$$$) — the bounds of the query.

It is guaranteed the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

Examples:
input:
3
7 5 3
1 2 3 2 1 2 3
1 7
2 7
3 7
8 4 2
4 3 1 1 2 4 3 2
3 6
1 5
5 4 2
4 5 1 2 3
1 4
1 5
output:
6
5
2
2
5
2
3

Note: In the second query of the first testcase, we calculate the following function values:

- $$$f([2,3,2,1,2])=3$$$ because Yunli can set $$$b_3=4$$$, $$$b_4=5$$$, and $$$b_5=6$$$, making a consecutive subarray of size $$$5$$$ in $$$3$$$ moves.
- $$$f([2,3,2,1,2,3])=2$$$ because we can set $$$b_3=0$$$ and $$$b_2=-1$$$, making a consecutive subarray of size $$$5$$$ in $$$2$$$ moves (starting at position $$$2$$$)

The answer to this query is $$$3+2=5$$$.

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

/**
 * with help
 * https://codeforces.com/problemset/problem/2009/G2
 */
public final class CF2009G2 implements Runnable{

    public static void main(String[] args) {
        new Thread(null, new CF2009G2(), "Main", 1<<28).start();
    }

    @Override
    public void run() {
        MyScanner sc = new MyScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
        // write here
        for(int t=sc.nextInt();t>0;t--){
            int n=sc.nextInt(),k=sc.nextInt(),q=sc.nextInt();
            int[]dict=new int[n];
            for(int i=0;i<dict.length;i++){
                dict[i]=sc.nextInt()-i;
            }
            Map<Integer,Integer>temp=new HashMap<>();
            TreeMap<Integer, Set<Integer>>count=new TreeMap<>();
            int[]costDict=new int[n-k+1];
            for(int i=0;i<dict.length;i++){
                temp.compute(dict[i], (key, o) -> {
                    if (o == null) {
                        count.computeIfAbsent(1, subKey -> new HashSet<>()).add(key);
                        return 1;
                    } else {
                        count.compute(o, (subKey, subO) -> {
                            if (subO.size() == 1) {
                                return null;
                            } else {
                                subO.remove(key);
                                return subO;
                            }
                        });
                        count.computeIfAbsent(o + 1, subKey -> new HashSet<>()).add(key);
                        return o + 1;
                    }
                });
                if(i>=k-1){
                    costDict[i-k+1]=k-count.lastKey();
                    temp.compute(dict[i-k+1], (key, o) -> {
                        count.compute(o, (subKey, subO) -> {
                            if (subO.size() == 1) {
                                return null;
                            } else {
                                subO.remove(key);
                                return subO;
                            }
                        });
                        if (o == 1) {
                            return null;
                        } else {
                            count.computeIfAbsent(o-1, subKey -> new HashSet<>()).add(key);
                            return o - 1;
                        }
                    });
                }
            }
            long[]res=new long[q];
            int[][]querys= IntStream.range(0,q).mapToObj(i->new int[]{sc.nextInt()-1,sc.nextInt()-1,i}).sorted((a,b)->b[0]-a[0]).toArray(int[][]::new);
            //{curValue,startIdex,lastIdx,preCostSum}
            Info[]mstack=new Info[n+1];
            mstack[0]=new Info(-1,n-k+1,n-k+1,0);
            int curIdx=n-k;
            int size=1;
            for(int[] query:querys){
                while (curIdx>=query[0]){
                    int startIdx=curIdx,endIdx=curIdx,preCostSum=costDict[curIdx];
                    while (mstack[size-1].curValue>=costDict[curIdx]){
                        endIdx=mstack[size-1].endIdx;
                        preCostSum+=1L*(mstack[size-1].endIdx-mstack[size-1].startIdx+1)*mstack[size-1].curValue;
                        size--;
                    }
                    preCostSum+=mstack[size-1].preCostSum;
                    mstack[size]=new Info(costDict[curIdx],startIdx,endIdx,preCostSum);
                    size++;
                    curIdx--;
                }
                res[query[2]]=mstack[size-1].preCostSum;
                int target=query[1]-k+1;
                int start=0;
                int end=size-1;
                while (start<=end){
                    int mid=(start+end)/2;
                    if(target>=mstack[mid].startIdx&&target<=mstack[mid].endIdx){
                        res[query[2]]-=(mstack[mid].preCostSum-1L*(target-mstack[mid].startIdx+1)*mstack[mid].curValue);
                        break;
                    }else if(target<mstack[mid].startIdx){
                        start=mid+1;
                    }else {
                        end=mid-1;
                    }
                }

            }
            for(int i=0;i<q;i++){
                out.println(res[i]);
            }
        }
        out.close();
    }

    private class Info{
        private int curValue;
        private int startIdx;
        private int endIdx;
        private long preCostSum;

        public Info(int curValue, int startIdx, int endIdx, long preCostSum) {
            this.curValue = curValue;
            this.startIdx = startIdx;
            this.endIdx = endIdx;
            this.preCostSum = preCostSum;
        }
    }

    public static class MyScanner {
        BufferedReader br;
        StringTokenizer st;

        public MyScanner() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

        String next() {
            while (st == null || !st.hasMoreElements()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

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

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

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

        String nextLine(){
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }
}

**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
1 2 3 4 5
1 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (hard version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $$$r \geq l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r \geq l+k-1$$$) — the bounds of the query.

It is guaranteed the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

Examples:
input:
3
7 5 3
1 2 3 2 1 2 3
1 7
2 7
3 7
8 4 2
4 3 1 1 2 4 3 2
3 6
1 5
5 4 2
4 5 1 2 3
1 4
1 5
output:
6
5
2
2
5
2
3

Note: In the second query of the first testcase, we calculate the following function values:

- $$$f([2,3,2,1,2])=3$$$ because Yunli can set $$$b_3=4$$$, $$$b_4=5$$$, and $$$b_5=6$$$, making a consecutive subarray of size $$$5$$$ in $$$3$$$ moves.
- $$$f([2,3,2,1,2,3])=2$$$ because we can set $$$b_3=0$$$ and $$$b_2=-1$$$, making a consecutive subarray of size $$$5$$$ in $$$2$$$ moves (starting at position $$$2$$$)

The answer to this query is $$$3+2=5$$$.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define endl '\n'
inline int read(){
	int x=0,k=1;char c=getchar();
	while(c<'0'||c>'9'){
		if(c=='-')k=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9')x=(x<<3)+(x<<1)+(c^48),c=getchar();
	return x*k;
}
const int N=5e5+5;
int a[N],b[N],c[N],f[N*4],d[N*4];
struct node{
	int l,r,i;
};
#define ul u<<1
#define ur u<<1|1
void puu(int u){
	f[u]=f[ul]+f[ur];
}
void build(int u,int l,int r){
	f[u]=0;
	if(l==r){
		return;
	}
	int m=l+r>>1;
	build(ul,l,m);
	build(ur,m+1,r);
}
void pp(int u,int l,int r,int x){
	f[u]=(r-l+1)*x;
	d[u]=x;
}
void pud(int u,int l,int r){
	if(!d[u])return;
	int m=l+r>>1;
	pp(ul,l,m,d[u]);
	pp(ur,m+1,r,d[u]);
	d[u]=0;
}
void up(int u,int l,int r,int L,int R,int x){
	if(L<=l&&r<=R){
		pp(u,l,r,x);
		return;
	}
	pud(u,l,r);
	int m=l+r>>1;
	if(m>=L)up(ul,l,m,L,R,x);
	if(m<R)up(ur,m+1,r,L,R,x);
	puu(u);
}
int qu(int u,int l,int r,int L,int R){
	if(L<=l&&r<=R){
		return f[u];
	}
	int re=0;
	pud(u,l,r);
	int m=l+r>>1;
	if(m>=L)re+=qu(ul,l,m,L,R);
	if(m<R)re+=qu(ur,m+1,r,L,R);
	return re;
}
void sol(){
	int n=read(),k=read(),m=read();
	for(int i=1;i<2*n;++i)b[i]=0;
	for(int i=1;i<=n;++i)a[i]=read()+n-i;
	priority_queue<pair<int,int>>q;
	for(int i=1;i<=n;++i){
		q.push({++b[a[i]],a[i]});
		if(i>=k){
			while(q.top().first!=b[q.top().second])q.pop();
			c[i]=k-q.top().first;
			q.push({--b[a[i-k+1]],a[i-k+1]});
		}
		else c[i]=0;
	}
	vector<vector<pair<int,int>>>v(n+1);
	vector<int>ans(m);
	for(int i=0;i<m;++i){
		int l=read(),r=read();
		v[l].push_back({r,i});
	}
	build(1,1,n);
	vector<int>s;
	for(int i=n;i;--i){
		int j=i+k-1;
		if(j>=k&&j<=n){
			while(!s.empty()&&c[*s.rbegin()]>c[j])s.pop_back();
			int r=n;
			if(!s.empty())r=*s.rbegin()-1;
			s.push_back(j);
			up(1,1,n,j,r,c[j]);
		}
		for(auto it:v[i]){
			ans[it.second]=qu(1,1,n,i+k-1,it.first);
		}
	}
	for(int i:ans)cout<<i<<endl;
}
signed main(){
	int T=read();
	while(T--)sol();
}

**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 given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to consider:

1. **Edge Cases for Array Length and Queries:**
   - Test with the minimum and maximum possible values for `n`, `k`, and `q`.
   - Test with `n` equal to `k` to see how the algorithm handles the smallest possible subarray length.

2. **Specific Values for Array Elements:**
   - Test with an array where all elements are the same.
   - Test with an array where elements are in increasing or decreasing order.
   - Test with an array that has a few distinct values repeated multiple times.

3. **Queries with Specific Ranges:**
   - Test queries where `l` and `r` are at the boundaries of the array.
   - Test queries where `r - l + 1` is exactly `k`.
   - Test queries where `r - l + 1` is much larger than `k`.

4. **Multiple Test Cases:**
   - Test with multiple test cases to ensure the algorithm can handle different scenarios.

Based on these considerations, here is a small-scale test input that should expose potential bugs in the buggy code:

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

This test input includes:
- A test case with an array of all the same elements.
- A test case with an array in increasing order.
- A test case with a larger array and larger subarray length.

These scenarios should help identify any issues in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (hard version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $$$r \geq l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r \geq l+k-1$$$) — the bounds of the query.

It is guaranteed the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

Examples:
input:
3
7 5 3
1 2 3 2 1 2 3
1 7
2 7
3 7
8 4 2
4 3 1 1 2 4 3 2
3 6
1 5
5 4 2
4 5 1 2 3
1 4
1 5
output:
6
5
2
2
5
2
3

Note: In the second query of the first testcase, we calculate the following function values:

- $$$f([2,3,2,1,2])=3$$$ because Yunli can set $$$b_3=4$$$, $$$b_4=5$$$, and $$$b_5=6$$$, making a consecutive subarray of size $$$5$$$ in $$$3$$$ moves.
- $$$f([2,3,2,1,2,3])=2$$$ because we can set $$$b_3=0$$$ and $$$b_2=-1$$$, making a consecutive subarray of size $$$5$$$ in $$$2$$$ moves (starting at position $$$2$$$)

The answer to this query is $$$3+2=5$$$.

Buggy Code:
/*
    Created by:- milind0110
*/
// #pragma GCC optimize("O3")
// #pragma GCC target ("avx2")
// #pragma GCC optimize("Ofast")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
// #pragma GCC optimize("unroll-loops")
#include<bits/stdc++.h>
// #pragma comment(linker, "/STACK:268435456");
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
using namespace std;
#define int long long
// #define ordered_set tree<pair<int,int>, null_type,less<pair<int,int>>, rb_tree_tag,tree_order_statistics_node_update>
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
const int inf = 1e12;
const int N = 2e5 + 5;
template<typename Node,typename Update>
class segtree{
    vector<Node> tree;
    vector<bool> lazy;
    vector<Update> upd;
    vector<Node> v;
    int32_t n,l,r;
    Update val;
    Node segnull;
    Update updnull;
    inline void buildTree(int32_t start,int32_t end,int32_t node){
        if(start == end){
            tree[node] = v[start];
            return ;
        }
        int mid = (start+end) >> 1;
        buildTree(start,mid,node << 1);
        buildTree(mid + 1,end,(node << 1) + 1);
        tree[node].merge(tree[node << 1],tree[(node << 1) + 1]);
    }
    inline void apply(int32_t start,int32_t end,int32_t node,Update update) {
        update.apply(tree[node],start,end);
        if(start != end) {
            upd[node << 1].combine(update,start,end);
            lazy[node << 1] = 1;
            upd[(node << 1) + 1].combine(update,start,end);
            lazy[(node << 1) + 1] = 1;
        }
    }
    inline void updateTree(int32_t start,int32_t end,int32_t node){
        if(start > end)
            return ;
        if(lazy[node]){
            apply(start,end,node,upd[node]);
            upd[node] = updnull;
            lazy[node] = 0;
        }
        if(l > end || r < start)
            return ;
        if(l <= start && end <= r){
            apply(start,end,node,val);
            return ;
        }
        int mid = (start+end) >> 1;
        updateTree(start,mid,node << 1);
        updateTree(mid + 1,end,(node << 1) + 1);
        tree[node].merge(tree[node << 1],tree[(node << 1) + 1]);
    }
    inline Node queryTree(int32_t start,int32_t end,int32_t node){
        if(lazy[node]){
            apply(start,end,node,upd[node]);
            upd[node] = updnull;
            lazy[node] = 0;
        }
        if(l > end || r < start)
            return segnull;
        if(l <= start && end <= r){
            return tree[node];
        }
        int mid = (start+end) >> 1;
        Node sol;
        Node left = queryTree(start,mid,node << 1);
        Node right = queryTree(mid + 1,end,(node << 1) + 1);
        sol.merge(left,right);
        return sol;
    }
public:
    template<typename T>
    segtree(int32_t n,T a,bool lazyflag = false) {
        //assign default values for Node and Update
        for(int i = 0; i < n; i++) v.push_back(a[i]);
        this->n = n;
        tree.resize(4 * n);
        lazy.resize(4 * n);
        if(lazyflag){
            upd.resize(4 * n);
        }
        buildTree(0,n - 1,1);
    }
    inline Node query(int32_t l,int32_t r) {
        this->l = l;
        this->r = r;
        return queryTree(0,n - 1,1);
    }
    inline void update(int32_t l,int32_t r,Update val) {
        this->val = val;
        this->l = l;
        this->r = r;
        updateTree(0,n - 1,1);
    }
    inline void update(int32_t ind,Update val) {
        this->val = val;
        this->l = ind;
        this->r = ind;
        updateTree(0,n - 1,1);
    }
};
class Node{
public:
	int x = 0;
	Node() {}
	Node(int x) {
        this->x = x;
	}
	void merge(Node &l,Node &r){
        x = l.x + r.x;
        //merge left and right
	}
};
class Update{
public:
	int x = inf;
	Update() {}
	Update(int x){
		this->x = x;
	}
	void combine(Update &parent,int32_t l,int32_t r){
        x = min(x,parent.x);
		//apply a parent update to a child update
	}
	void apply(Node &node,int32_t l,int32_t r){
        node.x = min(node.x,(r - l + 1) * x);
        	//apply an update to a range [l,r]
	}
};
void solve(){
    int n,k,q;
    cin >> n >> k >> q;
    vector<int> a(n);
    for(auto &x : a) cin >> x;
    map<int,int> cnt;
    set<pair<int,int>,greater<>> pq;
    vector<int> ans(n);
    for(int i = n - 1; i >= 0; i--){
        int x = a[i] - i;
        if(cnt.count(x)){
            pq.erase({cnt[x],x});
        }
        cnt[x]++;
        pq.insert({cnt[x],x});
        if(i + k < n){
            int x = a[i + k] - (i + k);
            pq.erase({cnt[x],x});
            cnt[x]--;
            if(cnt[x] == 0) cnt.erase(x);
            if(cnt.count(x)) pq.insert({cnt[x],x});
        }
        ans[i] = (k - (*pq.begin()).first);
    }
    segtree<Node,Update> st(n,ans,true);
    vector<vector<pair<int,int>>> qy(n);
    for(int i = 0; i < q; i++){
        int l,r;
        cin >> l >> r;
        l--;r--;
        qy[l].push_back({r,i});
    }
    vector<int> sol(q);
    for(int i = n - k; i >= 0; i--){
        st.update(i,n - 1,ans[i]);
        for(auto [r,ind] : qy[i]){
            sol[ind] = st.query(i,r - k + 1).x;
        }
    }
    for(int i = 0; i < q; i++) cout << sol[i] << "\n";
}
signed main() {
    auto begin = std::chrono::high_resolution_clock::now();
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    int32_t t = 1;
    cin >> t;
    for(int32_t i = 1; i <= t; i++){
        solve();
    }
    auto end = std::chrono::high_resolution_clock::now();
    auto elapsed = std::chrono::duration_cast<std::chrono::nanoseconds>(end - begin);
    cerr << "Time measured: " << elapsed.count() * 1e-9 << " seconds.\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
1
7 3 3
1 2 3 2 1 2 3
1 7
2 7
3 7
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (hard version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $$$r \geq l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r \geq l+k-1$$$) — the bounds of the query.

It is guaranteed the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

Examples:
input:
3
7 5 3
1 2 3 2 1 2 3
1 7
2 7
3 7
8 4 2
4 3 1 1 2 4 3 2
3 6
1 5
5 4 2
4 5 1 2 3
1 4
1 5
output:
6
5
2
2
5
2
3

Note: In the second query of the first testcase, we calculate the following function values:

- $$$f([2,3,2,1,2])=3$$$ because Yunli can set $$$b_3=4$$$, $$$b_4=5$$$, and $$$b_5=6$$$, making a consecutive subarray of size $$$5$$$ in $$$3$$$ moves.
- $$$f([2,3,2,1,2,3])=2$$$ because we can set $$$b_3=0$$$ and $$$b_2=-1$$$, making a consecutive subarray of size $$$5$$$ in $$$2$$$ moves (starting at position $$$2$$$)

The answer to this query is $$$3+2=5$$$.

Buggy Code:
import sys
import threading

def main():
    import sys
    import math
    t = int(sys.stdin.readline())
    total_n = 0
    total_q = 0
    for _ in range(t):
        n, k, q = map(int, sys.stdin.readline().split())
        total_n += n
        total_q += q
        a = list(map(int, sys.stdin.readline().split()))
        c = [a_i - (i + 1) for i, a_i in enumerate(a)]
        queries = [tuple(map(int, sys.stdin.readline().split())) for _ in range(q)]
        for l, r in queries:
            l -= 1  # Convert to 0-based indexing
            counts = {}
            max_freq = 0
            total = 0
            for j in range(l, r):
                c_j = c[j]
                counts[c_j] = counts.get(c_j, 0) + 1
                if counts[c_j] > max_freq:
                    max_freq = counts[c_j]
                if j >= l + k - 1:
                    f_j = k - max_freq
                    total += f_j
            print(total)
    # Ensure constraints are met
    assert total_n <= 2 * 10**5
    assert total_q <= 2 * 10**5

threading.Thread(target=main).start()

**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 consider edge cases and scenarios that maximize the failure probability. Here are the steps to think through the problem:

1. **Identify the Bug**: The buggy code seems to calculate the frequency of elements in a transformed array `c` and uses this to determine the minimum number of operations needed. However, it may not handle cases where the array has long sequences of the same value or where the transformation `c` has a large range of values.

2. **Edge Cases**:
   - **Long Sequences of the Same Value**: This can cause the frequency count to be incorrect.
   - **Large Range of Values in `c`**: This can cause the frequency dictionary to be very large and slow down the computation.
   - **Boundary Conditions**: Ensure the query ranges are at the boundaries of the array to test edge cases.

3. **Small-Scale Test Input**: We need a small-scale test input that is complete and valid, but challenging for the buggy code.

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

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

### Explanation:
- **Test Case 1**:
  - **Array Length (`n`)**: 5
  - **Subarray Length (`k`)**: 3
  - **Number of Queries (`q`)**: 2
  - **Array (`a`)**: [1, 1, 1, 1, 1]

- **Queries**:
  - **Query 1**: (1, 5)
  - **Query 2**: (2, 5)

### Analysis:
- **Query 1 (1, 5)**:
  - The subarrays to consider are:
    - [1, 1, 1]
    - [1, 1, 1, 1]
    - [1, 1, 1, 1, 1]
  - For each subarray, the minimum number of operations needed is 0 because there is already a consecutive subarray of length 3.
  - The total is \(0 + 0 + 0 = 0\).

- **Query 2 (2, 5)**:
  - The subarrays to consider are:
    - [1, 1, 1]
    - [1, 1, 1, 1]
  - For each subarray, the minimum number of operations needed is 0 because there is already a consecutive subarray of length 3.
  - The total is \(0 + 0 = 0\).

### Expected Output:
```plaintext
0
0
```

This test input is small and valid, but it should expose the bug in the buggy code if it incorrectly handles long sequences of the same value.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (hard version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $$$r \geq l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r \geq l+k-1$$$) — the bounds of the query.

It is guaranteed the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

Examples:
input:
3
7 5 3
1 2 3 2 1 2 3
1 7
2 7
3 7
8 4 2
4 3 1 1 2 4 3 2
3 6
1 5
5 4 2
4 5 1 2 3
1 4
1 5
output:
6
5
2
2
5
2
3

Note: In the second query of the first testcase, we calculate the following function values:

- $$$f([2,3,2,1,2])=3$$$ because Yunli can set $$$b_3=4$$$, $$$b_4=5$$$, and $$$b_5=6$$$, making a consecutive subarray of size $$$5$$$ in $$$3$$$ moves.
- $$$f([2,3,2,1,2,3])=2$$$ because we can set $$$b_3=0$$$ and $$$b_2=-1$$$, making a consecutive subarray of size $$$5$$$ in $$$2$$$ moves (starting at position $$$2$$$)

The answer to this query is $$$3+2=5$$$.

Buggy Code:
#!/usr/bin/env python
import os
import sys
from heapq import heapify, heappush, heappop
from collections import Counter
from io import BytesIO, IOBase

class RangeQuery:
    def __init__(self, data, func=min):
        self.func = func
        self._data = _data = [list(data)]
        i, n = 1, len(_data[0])
        while 2 * i <= n:
            prev = _data[-1]
            _data.append([func(prev[j], prev[j + i]) for j in range(n - 2 * i + 1)])
            i <<= 1

    def query(self, start, stop):
        """func of data[start, stop)"""
        depth = (stop - start).bit_length() - 1
        return self.func(self._data[depth][start], self._data[depth][stop - (1 << depth)])

    def binlift(self, start, lo):
        r = start
        for i in reversed(range(len(self._data))):
            if self._data[i][r] >= lo:
                r += 1 << i
        return r

    def __getitem__(self, idx):
        return self._data[0][idx]

def main():
    t = int(input())

    for _ in range(t):
        n, k, q = map(int, input().split())
        a = [int(ai) for ai in input().split()]

        cnt = [0] * (2 * n + 2)
        for i in range(k - 1):
            cnt[a[i] - i] += 1

        heap, removed = [-i for i in cnt if i], [0] * (2 * n + 2)
        heapify(heap)

        required = [0] * n
        for i in range(n - k + 1):
            if cnt[a[i + k - 1] - (i + k - 1)]:
                removed[cnt[a[i + k - 1] - (i + k - 1)]] += 1
            cnt[a[i + k - 1] - (i + k - 1)] += 1
            heappush(heap, -cnt[a[i + k - 1] - (i + k - 1)])

            if 0 <= i - 1:
                removed[cnt[a[i - 1] - (i - 1)]] += 1
                cnt[a[i - 1] - (i - 1)] -= 1
                if cnt[a[i - 1] - (i - 1)]:
                    heappush(heap, -cnt[a[i - 1] - (i - 1)])

            while removed[-heap[0]]:
                removed[-heappop(heap)] -= 1

            required[i] = k - -heap[0]

        rmq = RangeQuery(required)

        for _ in range(q):
            l, r = map(int, input().split())
            l -= 1
            r -= k - 1

            sol = 0
            full_lo = rmq.query(l, r)
            start, lo = l, required[l]
            while lo != full_lo:
                new_start = rmq.binlift(start, lo)
                sol += (new_start - start) * lo
                start, lo = new_start, required[new_start]

            sol += (r - start) * lo

            print(sol)

# region fastio

BUFSIZE = 8192

class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)

class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")

sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

# endregion

if __name__ == "__main__":
    main()

**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 2
1 2 3 4 5
1 5
2 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Yunli's Subarray Queries (hard version)
time_limit_ms: 3000
memory_limit_mb: 512
Description: This is the hard version of the problem. In this version, it is guaranteed that $$$r \geq l+k-1$$$ for all queries.

For an arbitrary array $$$b$$$, Yunli can perform the following operation any number of times:

- Select an index $$$i$$$. Set $$$b_i = x$$$ where $$$x$$$ is any integer she desires ($$$x$$$ is not limited to the interval $$$[1,n]$$$).

Denote $$$f(b)$$$ as the minimum number of operations she needs to perform until there exists a consecutive subarray$$$^{\text{∗}}$$$ of length at least $$$k$$$ in $$$b$$$.

Yunli is given an array $$$a$$$ of size $$$n$$$ and asks you $$$q$$$ queries. In each query, you must output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$.

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

The first line of each test case contains three integers $$$n$$$, $$$k$$$, and $$$q$$$ ($$$1 \leq k \leq n \leq 2 \cdot 10^5$$$, $$$1 \leq q \leq 2 \cdot 10^5$$$) — the length of the array, the length of the consecutive subarray, and the number of queries.

The following line contains $$$n$$$ integers $$$a_1, a_2, ..., a_n$$$ ($$$1 \leq a_i \leq n$$$).

The following $$$q$$$ lines contain two integers $$$l$$$ and $$$r$$$ ($$$1 \leq l \leq r \leq n$$$, $$$r \geq l+k-1$$$) — the bounds of the query.

It is guaranteed the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$ and the sum of $$$q$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$.

Output: Output $$$\sum_{j=l+k-1}^{r} f([a_l, a_{l+1}, \ldots, a_j])$$$ for each query on a new line.

Examples:
input:
3
7 5 3
1 2 3 2 1 2 3
1 7
2 7
3 7
8 4 2
4 3 1 1 2 4 3 2
3 6
1 5
5 4 2
4 5 1 2 3
1 4
1 5
output:
6
5
2
2
5
2
3

Note: In the second query of the first testcase, we calculate the following function values:

- $$$f([2,3,2,1,2])=3$$$ because Yunli can set $$$b_3=4$$$, $$$b_4=5$$$, and $$$b_5=6$$$, making a consecutive subarray of size $$$5$$$ in $$$3$$$ moves.
- $$$f([2,3,2,1,2,3])=2$$$ because we can set $$$b_3=0$$$ and $$$b_2=-1$$$, making a consecutive subarray of size $$$5$$$ in $$$2$$$ moves (starting at position $$$2$$$)

The answer to this query is $$$3+2=5$$$.

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

public class Main {
    public static int MDC = 998244353;

    public static void main(String[] args) throws IOException {
        sc = new MScanner(System.in);
        pw = new PrintWriter(System.out);
        int t = sc.nextInt();
        //int t = 1;
        for (int i = 0; i < t; i++) {
            solver();
        }
    }

    static long[] bitArr;

    public static void solver() throws IOException {
        int n = sc.nextInt();
        int k = sc.nextInt();
        int q = sc.nextInt();
        int[] a = new int[n];
        int[] b = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
            b[i] = a[i] - i;
        }
        int[] la = new int[n];
        int r = 0;
        int max = 0;
        Map<Integer, Set<Integer>> map = new HashMap<>();
        Map<Integer, Integer> map1 = new HashMap<>();
        for (int i = 0; i < n; i++) {
            while (r < n && r < i + k) {
                if (map1.containsKey(b[r])) {
                    Integer i1 = map1.get(b[r]);
                    // 移除原来的元素
                    Set<Integer> integers = map.get(i1);
                    integers.remove(b[r]);
                    map.put(i1, integers);
                    // 更新元素
                    map1.put(b[r], i1 + 1);
                    Set<Integer> orDefault = map.getOrDefault(i1 + 1, new HashSet<>());
                    orDefault.add(b[r]);
                    map.put(i1 + 1, orDefault);
                } else {
                    // 更新元素
                    map1.put(b[r], 1);
                    Set<Integer> orDefault = map.getOrDefault(1, new HashSet<>());
                    orDefault.add(b[r]);
                    map.put(1, orDefault);
                }
                max = Math.max(max, map1.get(b[r]));
                r++;
            }
            la[i] = k - max;
            // 删除元素更新最大值
            Integer i1 = map1.get(b[i]);
            map1.put(b[i], i1 - 1);

            Set<Integer> integers = map.get(i1);
            integers.remove(b[i]);
            map.put(i1, integers);

            Set<Integer> integers1 = map.getOrDefault(i1 - 1, new HashSet<>());
            integers1.add(b[i]);
            map.put(i1 - 1, integers1);
            if (i1 == max && map.get(i1).isEmpty()) {
                max--;
            }
        }
        // 目前计算的为la数组
        List<int[]> list = new ArrayList<>();
        for (int i = 0; i < q; i++) {
            int ql = sc.nextInt();
            int qr = sc.nextInt();
            list.add(new int[]{ql, qr, i});
        }
        list.sort((o1, o2) -> o1[0] == o2[0] ? o1[1] - o2[1] : o2[0] - o1[0]);
        TreeSet<long[]> treeSet = new TreeSet<>((o1, o2) -> Long.compare(o1[0], o2[0]));
        int p = n - 1;
        long[] nn = new long[q];
        for (int i = 0; i < list.size(); i++) {
            int nl = list.get(i)[0] - 1;
            int nr = list.get(i)[1] - 1;
            while (p >= nl) {
                long num = 0;
                while (!treeSet.isEmpty() && treeSet.first()[1] >= la[p]) {
                    long[] ints = treeSet.pollFirst();
                    num++;
                }
                num = (num + 1) * la[p];
                if (!treeSet.isEmpty()) num += treeSet.first()[2];
                treeSet.add(new long[]{p, la[p], num});
                p--;
            }
            // 计算值
            long[] lower = treeSet.lower(new long[]{nr - k + 2, -1, -1});

            nn[list.get(i)[2]] = treeSet.first()[2]-lower[2]+lower[1]*(nr-k +1-lower[0]+1);
        }
        for (int i = 0; i < q; i++) {
            pw.println(nn[i]);
        }
        pw.flush();
    }

    public static void bitInsert(int index, long v, int n) {
        while (index <= n) {
            bitArr[index] += v;
            index += (index & (-index));
        }
    }

    public static long bitQuery(int index, int n) {
        long ans = 0;
        while (index > 0) {
            ans += bitArr[index];
            index -= -index & index;
        }
        return ans;
    }

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

    private static long pow(long n, long m) {
        long res = 1;
        long base = n;
        while (m != 0) {
            if ((m & 1) == 1) {
                res = res * base;
                res %= MDC;
            }
            base = base * base;
            base %= MDC;
            m = m >> 1;
        }
        return res;
    }

    static int curt;
    static PrintWriter pw;
    static MScanner sc;

    static class MScanner {
        StringTokenizer st;
        BufferedReader br;

        public MScanner(InputStream system) {
            br = new BufferedReader(new InputStreamReader(system));
        }

        public MScanner(String file) throws Exception {
            br = new BufferedReader(new FileReader(file));
        }

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

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

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

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

        public char nextChar() throws IOException {
            return next().charAt(0);
        }

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

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

        public void waitForInput() throws InterruptedException {
            Thread.sleep(3000);
        }

    }

    static void dbg(int[] in) {
        System.out.println(Arrays.toString(in));
    }

    static void dbg(long[] in) {
        System.out.println(Arrays.toString(in));
    }

    static void sort(int[] in) {
        shuffle(in);
        Arrays.sort(in);
    }

    static void sort(long[] in) {
        shuffle(in);
        Arrays.sort(in);
    }

    static void shuffle(int[] in) {
        for (int i = 0; i < in.length; i++) {
            int idx = (int) (Math.random() * in.length);
            int tmp = in[i];
            in[i] = in[idx];
            in[idx] = tmp;
        }
    }

    static void shuffle(long[] in) {
        for (int i = 0; i < in.length; i++) {
            int idx = (int) (Math.random() * in.length);
            long tmp = in[i];
            in[i] = in[idx];
            in[idx] = tmp;
        }
    }
}

class SegmentTree {

    TreeNode[] tree;

    private class TreeNode {
        int left, right;
        int sum, lazy;

        public TreeNode(int left, int right) {
            this.left = left;
            this.right = right;
        }

        public int mid() {
            return left + (right - left) / 2;
        }

        public int length() {
            return right - left + 1;
        }
    }

    public SegmentTree(int[] nums) {
        tree = new TreeNode[4 * nums.length];
        build(0, 0, nums.length - 1, nums);
    }

    public void build(int index, int start, int end, int[] nums) {
        tree[index] = new TreeNode(start, end);
        if (start == end) {
            tree[index].sum = nums[start];
            return;
        }
        int mid = tree[index].mid();
        build(2 * index + 1, start, mid, nums);
        build(2 * index + 2, mid + 1, end, nums);
        pushUp(index);
    }

    public int search(int index, int start, int end) {
        if (tree[index].left == start && tree[index].right == end) {
            return tree[index].sum;
        }
        pushDown(index);
        int mid = tree[index].mid(), left = 2 * index + 1, right = 2 * index + 2;
        int sum = 0;
        if (mid >= end) {
            sum += search(left, start, end);
        } else if (mid < start) {
            sum += search(right, start, end);
        } else {
            sum += search(left, start, mid);
            sum += search(right, mid + 1, end);
        }
        return sum;
    }

    public int search1(int index, int start, int end) {
        if (tree[index].left >= start && tree[index].right <= end) {
            return tree[index].sum;
        }
        pushDown(index);
        int sum = 0, mid = tree[index].mid();
        if (mid >= start) {
            sum += search1(2 * index + 1, start, end);
        }
        if (mid < end) {
            sum += search1(2 * index + 2, start, end);
        }
        return sum;
    }

    public void update(int index, int start, int end, int value) {
        if (tree[index].left == start && tree[index].right == end) {
            tree[index].lazy += value;
            tree[index].sum += value * (tree[index].length());
            return;
        }
        if (tree[index].left == tree[index].right) {
            return;
        }
        pushDown(index);
        int mid = tree[index].mid(), left = 2 * index + 1, right = 2 * index + 2;
        if (mid >= end) {
            update(left, start, end, value);
        } else if (mid < start) {
            update(right, start, end, value);
        } else {
            update(left, start, mid, value);
            update(right, mid + 1, end, value);
        }
        pushUp(index);
    }

    public void update1(int index, int start, int end, int value) {
        if (tree[index].left >= start && tree[index].right <= end) {
            tree[index].lazy += value;
            tree[index].sum += value * tree[index].length();
            return;
        }
        pushDown(index);
        int mid = tree[index].mid();
        if (mid >= start) {
            update1(2 * index + 1, start, end, value);
        }
        if (mid < end) {
            update1(2 * index + 2, start, end, value);
        }
        pushUp(index);
    }

    public void pushDown(int index) {
        if (tree[index].lazy != 0) {
            int lazy = tree[index].lazy;
            int left = 2 * index + 1;
            int right = 2 * index + 2;
            tree[left].lazy += lazy;
            tree[right].lazy += lazy;
            tree[left].sum += lazy * tree[left].length();
            tree[right].sum += lazy * tree[right].length();
            tree[index].lazy = 0;
        }
    }

    public void pushUp(int index) {
        tree[index].sum = tree[2 * index + 1].sum + tree[2 * index + 2].sum;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to create such a test input:

1. **Small Scale**: Ensure the test case is small enough to be manageable but still challenging.
2. **Edge Cases**: Include edge cases such as:
   - Queries with the minimum and maximum possible values.
   - Arrays with all elements being the same.
   - Arrays with a single element.
   - Queries that span the entire array.
3. **Specific Conditions**: Ensure that the conditions \( r \geq l + k - 1 \) are met for all queries.

Given these considerations, here is a small-scale test input that should expose potential bugs in the code:

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

### Explanation:
1. **Test Case 1**:
   - Array: `[1, 2, 3, 4, 5]`
   - Query 1: `(1, 5)` - This spans the entire array.
   - Query 2: `(2, 5)` - This spans a part of the array starting from the second element.

2. **Test Case 2**:
   - Array: `[1, 2, 3, 4, 5, 6, 7]`
   - Query 1: `(1, 7)` - This spans the entire array.
   - Query 2: `(2, 7)` - This spans a part of the array starting from the second element.
   - Query 3: `(3, 7)` - This spans a part of the array starting from the third element.

These test cases are designed to:
- Test the code's ability to handle queries that span the entire array.
- Test the code's ability to handle queries that start from different positions within the array.
- Test the code's ability to handle arrays with a small number of elements and a larger number of elements.

By focusing on these scenarios, we can effectively expose any bugs in the provided buggy code.