**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

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

public class Main {
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    static BufferedWriter writer = new BufferedWriter(new OutputStreamWriter(System.out));

    public static void main(String[] args) throws IOException {
        int t = 1;
//        t = Integer.parseInt(reader.readLine());
        while (--t >= 0) solve();
        writer.flush();
        writer.close();
    }

    static void solve() throws IOException {
        String[] s = reader.readLine().split(" ");
        int n = Integer.parseInt(s[0]), q = Integer.parseInt(s[1]);
        int[] a = new int[n];
        s = reader.readLine().split(" ");
        for (int i = 0; i < n; i++) a[i] = Integer.parseInt(s[i]);
        List<Integer>[] h = new List[200001]; // height -> list of indices higher than height
        for (int i = 0; i < 200001; i++) h[i] = new ArrayList<>();
        for (int i = 0; i < n; i++) for (int j = 1; j <= a[i]; j++) h[j].add(i);
        while (--q >= 0) {
            s = reader.readLine().split(" ");
            int tar = Integer.parseInt(s[0]) - 1, x = Integer.parseInt(s[1]);
            int level = 1, tarLevel = a[tar], l = 0;
            for (; level <= tarLevel; level++) {
                int idx = bs(h[level], l);
                if (idx + x - 1 >= h[level].size()) {
                    if (level >= tarLevel) l = h[level].get(h[level].size() - 1) + 1;
                    break;
                }
                l = h[level].get(idx + x - 1) + 1;
            }
            if (l <= tar) writer.write("NO\n");
            else writer.write("YES\n");
        }
    }

    static int bs(List<Integer> list, int target) {
        int l = 0, r = list.size() - 1;
        while (l <= r) {
            int m = l + (r - l) / 2;
            if (list.get(m) < target) l = m + 1;
            else r = m - 1;
        }
        return l;
    }
}

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

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

Think step by step.

Response:
```plaintext
4 4
2 1 2 1
1 1
2 1
3 1
4 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

Buggy Code:
#include <cstdio>
#include <vector>
#include <numeric>
#include <stack>
#include <cassert>
using namespace std;

class SegmentTree {
	public:
	int n;
	vector<int> tree;
	int size() {
		return n;
	}

	SegmentTree(int n) {
		this-> n = n;
		tree = vector<int>(4 * n + 4, 0);
	}

	void inc_pos(int pos, int val) {
		int cur = 1;
		int left = 0;
		int right = n;
		while(left <= right) {
			tree[cur] += val;
			//printf("inc %d %d\n", left, right);
			if (left == right){
				break;
			}
			int mid = (left + right) / 2;
			if(pos <= mid) {
				cur = cur * 2;
				right = mid;
			} else {
				cur =  2 * cur + 1;
				left = mid + 1;
			}
		}
	}

	int get(int start, int end) {
		int ans = 0;
		stack< vector<int> > node_front;
		node_front.push(vector<int>({1, 0, n}));

		while(!node_front.empty()) {
			int cur = node_front.top()[0];
			int left = node_front.top()[1];
			int right = node_front.top()[2];
			node_front.pop();
			if(left >= start && right <= end) {
				ans += tree[cur];
				//printf("add %d %d %d %d\n", left, right, tree[cur], ans);
			} else {
				int mid = (left + right) / 2;
				if (mid >= start) {
					node_front.push(vector<int>({cur * 2, left, mid}));
				}
				if(mid + 1 <= end) {
					node_front.push(vector<int>({cur * 2 + 1, mid + 1, right}));
				}
			}
		}
		return ans;
	}

};
int test_segTree() {
	SegmentTree segTree(15);
	segTree.inc_pos(1, 1);
	segTree.inc_pos(2, 1);
	assert(segTree.get(1, 15) == 2);
	assert(segTree.get(2, 15) == 1);
	return 0;
}

int main() {
	//test_segTree();
	//return 0;
	int n;
	int Q;
	scanf("%d%d", &n, &Q);

	vector<int> arr(n);
	for(int i=0;i < n;i++){
		scanf("%d", &arr[i]);
	}

	SegmentTree segTree(n);
	vector<int> min_k_to_fight(n);
	int cur_k;
	for(int i = 0; i < n; i++) {
		int high = i + 1, low = 1;
		while(low < high) {
			int mid = (low + high) / 2;
			int prev_fights =  segTree.get(1, mid);
			bool will_fight = (prev_fights/mid + 1) <= arr[i];
			if (will_fight) {
				high = mid;
			} else {
				low = mid +1;
			}
		}
		min_k_to_fight[i] = low;
		//printf("i = %d min_k = %d\n", i, low);
		segTree.inc_pos(low, 1);
	}

	for (int i = 0; i < Q; i++) {
		int I, X;
		scanf("%d%d", &I, &X);
		if(X >= min_k_to_fight[I-1]) {
			printf("YES\n");
		} else {
			printf("NO\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:
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 can cause the segment tree or the binary search logic to fail. Here are the steps to think through this:

1. **Edge Cases for Segment Tree**:
   - Ensure the segment tree handles the boundary conditions correctly.
   - Test with a small number of elements to see if the segment tree updates and queries work as expected.

2. **Edge Cases for Binary Search**:
   - Test with a sequence where the levels of monsters are not strictly increasing or decreasing.
   - Test with a sequence where the levels of monsters are the same or very close to each other.
   - Test with a sequence where the levels of monsters are much higher than the initial level of Monocarp.

3. **Queries**:
   - Test with queries where the value of `k` is very small or very large.
   - Test with queries where the value of `k` is exactly the minimum value required to fight a monster.
   - Test with queries where the value of `k` is just below the minimum value required to fight a monster.

Given these considerations, a small-scale test input that can expose the bug might look like this:

```plaintext
5 10
2 2 3 3 4
1 1
2 1
3 1
4 1
5 1
1 2
2 2
3 2
4 2
5 2
```

This test input includes:
- A sequence of 5 monsters with levels that are not strictly increasing.
- 10 queries to test various values of `k` for each monster.

Let's break down the expected output for this test input:

1. For `k = 1`:
   - Monocarp will fight all monsters because the level up happens after each fight.
   - Expected output: `YES YES YES YES YES`

2. For `k = 2`:
   - Monocarp will fight the first monster and then level up.
   - Monocarp will flee the second monster because his level is 2 and the monster's level is 2.
   - Monocarp will fight the third monster and then level up.
   - Monocarp will flee the fourth monster because his level is 3 and the monster's level is 3.
   - Monocarp will fight the fifth monster and then level up.
   - Expected output: `YES NO YES NO YES`

So, the complete expected output for the test input is:

```plaintext
YES
YES
YES
YES
YES
YES
NO
YES
NO
YES
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

Buggy Code:
#include "bits/stdc++.h"

using namespace std;

using ll = long long;
using ld = long double;

const ld PI = 3.141592653589793;
const int MOD = 1e9+7;
const int INF = 1e9;
const ll INFLL = 4e18;
const double EPS = 1e-9;

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... x) {__print(t); if (sizeof...(x)) cerr << ", "; _print(x...);}

#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#ifdef ONLINE_JUDGE
    #define debug(x...) 42
#endif

const int MAXN = 2e5 + 1;
const int LIM = 480; // >= LIM
const int ALIM = MAXN / LIM + 3;
int pref[ALIM + 2][MAXN];

void solve(int NT) {
    debug(MAXN, LIM, ALIM);
    int n, q; cin >> n >> q;

    vector<int> a(n);
    for(int i = 0; i < n; ++i) {
        cin >> a[i];

        int to = min(ALIM, a[i]);
        for(int j = to; j >= 1; --j) pref[j][i]++;
    }

    for(int i = 1; i <= ALIM; ++i) {
        for(int j = 0; j < n; ++j) {
            pref[i][j] += pref[i][j - 1];
        }
        pref[i][n] = INF;
    }

    vector<int> ans(q);
    vector<vector<pair<int, int>>> que(n + 1);
    for(int i = 0; i < q; ++i) {
        int x, id; cin >> id >> x; id--;
        que[x].emplace_back(-id, i);
    }

    for(int i = 1; i <= n; ++i) {
        sort(que[i].begin(), que[i].end());
    }

    // for(int i = 0; i <= 100; ++i) {
    //     cerr << pref[1][i] << ' ';
    // }
    // cerr << '\n';

    for(int k = 1; k <= n; ++k) {
        if(k <= LIM) {
            int cnt = 0, cur = 1;
            for(int i = 0; i < n; ++i) {
                bool ok = false;
                if(cur <= a[i]) {
                    ok = true;
                    cnt++;
                    if(cnt % k == 0) {
                        cur++;
                    }
                }

                // debug(k, i, ok);
                while(!que[k].empty() && que[k].back().first == -i) {
                    ans[que[k].back().second] = ok;
                    que[k].pop_back();
                }
            }

        } else {
            int cnt = 0, cur = 1, last = 0;

            while(last < n) {
                // debug(k, cnt, cur, last);
                int i = lower_bound(pref[cur] + last, pref[cur] + n, cnt+k) - pref[cur];
                // [last, i]
                // debug(i);
                while(!que[k].empty() && (-que[k].back().first) <= i) {
                    if(a[-que[k].back().first] >= cur) {
                        ans[que[k].back().second] = 1;
                    }
                    que[k].pop_back();
                }
                cnt += k;
                cur++;
                last = i + 1;
            }

        }
        que[k].clear();
        que[k].shrink_to_fit();
    }

    for(int x: ans) {
        if(x == 1) cout << "YES\n";
        else cout << "NO\n";
    }
}

// dp1[x][v] = AND dp2[u][v-a[x]]
// dp2[x][v] =

// #define TESTCASES
int main() {
    cin.tie(0)->sync_with_stdio(0);

    int t = 1;
    #ifdef TESTCASES
        cin >> t;
    #endif

    for(int i = 1; i <= t; ++i){
        solve(i);
        cout << "\n";
    }
}

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

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

Think step by step.

Response:
```plaintext
5 10
3 2 1 2 3
1 1
2 1
3 1
4 1
5 1
1 2
2 2
3 2
4 2
5 2
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

Buggy Code:
t = input().split()
n = int(t[0])
q = int(t[1])
arr = input().split()
for _ in range(q):
    s = input().split()
    i = int(s[0])
    k = int(s[1])
    level = 1
    fight = 0
    for j in range(1, i):
        if int(arr[j-1]) < level:
            continue
        fight += 1
        if fight%k == 0:
            level += 1
    if int(arr[i-1]) < level:
        print("NO")
    else:
        print("YES")

**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 given buggy code, we need to focus on edge cases and scenarios that are likely to cause the code to fail. The main issues in the buggy code are:

1. The code does not correctly handle the level-up logic, especially when the number of fights required for a level up (`k`) is large.
2. The code does not efficiently handle the case where the number of monsters (`n`) is large, which can lead to performance issues.

Given these points, we can create a test case that:

1. Has a large number of monsters with levels that are just above or below Monocarp's level.
2. Includes queries where the number of fights required for a level up (`k`) is large, which can cause the code to fail due to incorrect level-up logic.

Here is a small-scale test input that should expose the bug:

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

### Explanation:
- **Number of monsters (`n`):** 10
- **Number of queries (`q`):** 5
- **Levels of monsters:** 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
- **Queries:**
  - `1 1`: Monocarp starts at level 1 and needs 1 fight to level up. The first monster has level 1, so Monocarp will fight it.
  - `5 2`: Monocarp starts at level 1 and needs 2 fights to level up. By the time he reaches the 5th monster, he will have fought 4 monsters and will be at level 2. The 5th monster has level 5, so Monocarp will fight it.
  - `8 3`: Monocarp starts at level 1 and needs 3 fights to level up. By the time he reaches the 8th monster, he will have fought 7 monsters and will be at level 3. The 8th monster has level 8, so Monocarp will fight it.
  - `10 4`: Monocarp starts at level 1 and needs 4 fights to level up. By the time he reaches the 10th monster, he will have fought 9 monsters and will be at level 3. The 10th monster has level 10, so Monocarp will fight it.
  - `10 5`: Monocarp starts at level 1 and needs 5 fights to level up. By the time he reaches the 10th monster, he will have fought 9 monsters and will be at level 2. The 10th monster has level 10, so Monocarp will fight it.

This test input should expose the bug in the level-up logic and the handling of large values of `k`.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

Buggy Code:
def process_queries(n, q, monster_levels, queries):
    results = []
    cache = {}

    for i, k in queries:
        if k not in cache:
            cache[k] = {}  # Создаем новую таблицу для этого значения k

        # Инициализация уровня Монокарпа и количества сражений для текущего запроса
        if i in cache[k]:
            level_monocarp, fights_count = cache[k][i]
        else:
            level_monocarp = 1
            fights_count = 0

            # Сражаемся с монстрами от 1 до i
            for j in range(1, i):
                if level_monocarp <= monster_levels[j - 1]:
                    fights_count += 1
                    if fights_count % k == 0:
                        level_monocarp += 1

            # Сохраняем состояние после обработки монстров до индекса i
            cache[k][i] = (level_monocarp, fights_count)

        # Проверяем, сможет ли Монокарп победить i-го монстра
        if level_monocarp > monster_levels[i - 1]:
            results.append("NO")
        else:
            results.append("YES")

    return results

def main():
    import sys
    input = sys.stdin.read
    data = input().split()

    # Чтение данных
    index = 0

    n = int(data[index])
    index += 1
    q = int(data[index])
    index += 1

    monster_levels = [int(data[index + i]) for i in range(n)]
    index += n

    queries = []
    for _ in range(q):
        i = int(data[index])
        k = int(data[index + 1])
        queries.append((i, k))
        index += 2

    # Обработка запросов
    results = process_queries(n, q, monster_levels, queries)

    # Вывод результатов
    for result in results:
        print(result)

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:
To create a challenging test input that exposes the bug in the given code, we need to focus on edge cases and scenarios that are likely to cause the code to fail. The code has a caching mechanism, which might not handle certain cases correctly, especially when the number of fights required to level up (k) is very small or very large.

### Step-by-Step Reasoning:

1. **Small k Values**: When k is very small, the number of fights required to level up is minimal. This can lead to rapid level increases, which might not be correctly tracked by the cache.
2. **Large k Values**: When k is very large, the level might not increase at all, or only after a large number of fights. This can also expose issues in the cache.
3. **Monsters with Levels Close to Monocarp's Initial Level**: If the monsters' levels are close to Monocarp's initial level, it can be challenging to determine if he will fight or flee.
4. **Monsters with Levels Much Higher than Monocarp's Initial Level**: This can also be a challenging scenario, especially if k is large and Monocarp's level increases slowly.

### Chosen Test Input:
Let's create a test input with:
- A small number of monsters (n = 5) to keep the input small.
- A mix of monster levels to test different scenarios.
- A mix of k values to test both small and large k.
- Queries that cover a range of indices and k values.

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

### Explanation:
- **Monsters' Levels**: [1, 2, 3, 4, 5]
- **Queries**:
  - (1, 1): Monocarp's level is 1, and k is 1. He will fight the first monster.
  - (2, 2): Monocarp's level is 1, and k is 2. He will fight the second monster.
  - (3, 3): Monocarp's level is 1, and k is 3. He will fight the third monster.
  - (4, 4): Monocarp's level is 1, and k is 4. He will fight the fourth monster.
  - (5, 5): Monocarp's level is 1, and k is 5. He will fight the fifth monster.
  - (1, 5): Monocarp's level is 1, and k is 5. He will fight the first monster.
  - (3, 1): Monocarp's level is 1, and k is 1. He will fight the third monster.
  - (5, 1): Monocarp's level is 1, and k is 1. He will fight the fifth monster.

This test input covers a variety of scenarios and should help expose any issues in the caching mechanism or the logic for determining whether Monocarp fights or flees.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

Buggy Code:
import java.util.*;

public class task_168_E_v3 { //version 3 uses HashMap

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = Integer.parseInt(scanner.next());
        int q = Integer.parseInt(scanner.next());
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(scanner.next());
        }
        Map[] levels = new Map[n + 1];

        for (int j = 0; j < q; j++) {
            int i = Integer.parseInt(scanner.next()) - 1;
            int x = Integer.parseInt(scanner.next());
            System.out.println(ans(i, x, a, levels[x]));
        }
    }

    private static String ans(int i, int x, int[] a, Map map) {
        int level = 1;
        if (map == null) {
            map = new HashMap();
            map.put(0, -1);
        }
        int lastLevel = map.size();
        if (lastLevel == 0) {
            level = findLevel(0, i, x, a, map);
        } else {
            int lastUpNumber = (Integer) map.get(lastLevel - 1);
            if (i == lastUpNumber) {
                return "YES";
            }
            if (i < lastUpNumber) {
                if (lastLevel <= a[i]) {
                    return "YES";
                } else {
                    //find in Levels step for the a[i] level
                    int step = (Integer) map.get(a[i] + 1);
                    return i > step ? "NO" : "YES";
                }
            } else {
                level = findLevel(lastUpNumber + 1, i, x, a, map);

            }
        }
        return a[i] < level ? "NO" : "YES";
    }

    private static int findLevel(int from, int to, int x, int[] a, Map map) {
        int level = map.size();
        int killed = 0;
        for (int i = from; i < to; i++) {
            if (a[i] >= level) {
                killed++;
            }
            if (killed == x) {
                level++;
                map.put(level, i);
                killed = 0;
            }
        }
        return level;
    }
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

Buggy Code:
import java.util.*;

public class task_168_E_v2 {

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int n = Integer.parseInt(scanner.next());
        int q = Integer.parseInt(scanner.next());
        int[] a = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = Integer.parseInt(scanner.next());
        }
        List[] levels = new List[n + 1];
        HashMap<Integer, Integer> checked = new HashMap<>();

        for (int j = 0; j < q; j++) {
            int i = Integer.parseInt(scanner.next()) - 1;
            int x = Integer.parseInt(scanner.next());
            System.out.println(ans(i, x, a, levels));
        }
    }

    private static String ans(int i, int x, int[] a, List[] levels) {
        int level = 1;
        List list = levels[x];
        if (list == null) {
            list = new ArrayList();
            list.add(-1);
        }
        int lastLevel = list.size();
        if (lastLevel == 0) {
            level = findLevel(0, i, x, a, list);
        } else {
            int lastUpNumber = (Integer) list.get(lastLevel - 1);
            if (i == lastUpNumber) {
                return "YES";
            }
            if (i < lastUpNumber) {
                if (lastLevel <= a[i]) {
                    return "YES";
                } else {
                    //find in Levels
                    level = Collections.binarySearch(list, i); // maybe +1/-1 ???
                }
            } else {
                level = findLevel(lastUpNumber + 1, i, x, a, list);

            }
        }
        return a[i] < level ? "NO" : "YES";
    }

    private static int findLevel(int from, int to, int x, int[] a, List list) {
        int level = list.size();
        int killed = 0;
        for (int i = from; i < to; i++) {
            if (a[i] >= level) {
                killed++;
            }
            if (killed == x) {
                level++;
                list.add(i);
                killed = 0;
            }
        }
        return level;
    }
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

Buggy Code:
# ﷽
import sys

from math import ceil,sqrt

input = lambda: sys.stdin.buffer.readline().decode().rstrip()

from itertools import accumulate

def solve():
    # n = int(input())
    n,q = list(map(int,input().split()))

    arr =  list(map(int,input().split()))
    # arr = [5000 for i in range(50000)]
    # n = len(arr)
    sq = 2*ceil(sqrt(n)) + 1

    ans = ["NO"]*q
    less = {}
    more = {}

    for i in range(q) :
        j,x = list(map(int,input().split()))
        if x <= sq :
            try :
                less[(j,x)].append(i)
            except :
                less[j,x] = [i]
        else:
            try :
                more[(j,x)].append(i)
            except :
                more[j,x] = [i]

    for k in range(sq+1) :
        cur = 1
        fought = 0

        for i in range(n) :
            if arr[i] >= cur :
                fought += 1
                if fought == k :
                    fought = 0
                    cur += 1

                while (i+1,k) in less and less[i+1,k]:
                    j = less[i+1,k].pop()
                    ans[j] = "YES"

    grid = [[0  for j in range(n+1)] for i in range(sq+1)]

    for i in range(n) :
        ai = min(arr[i],sq)
        grid[ai][i+1] += 1

    for ai in range(sq+1) :
        grid[ai] = list(accumulate(grid[ai]))

    for j in range(1,sq+1) :
        for i in range(n+1) :
            grid[j][i] += grid[j-1][i]

    # print(more)
    # print(ceil(sqrt(2*10**5))*2*10**5)
    for j,x in more :
        cur = x
        # print(j,x)
        if j <= x :
            while more[j,x] :
                i = more[j,x].pop()
                ans[i] = "YES"
        else:
            val = arr[j-1]
            cur = x
            for lvl in range(2,(n+x-1)//x +1) :
                l = cur + 1
                r = n
                g = 0
                while r > l :
                    m = (r+l)>>1
                    if (m-cur) - (grid[lvl-1][m] - grid[lvl-1][cur-1]) >= x :
                        r = m
                    else:
                        l = m + 1

                # if (r-cur) - (grid[lvl-1][r] - grid[lvl-1][cur-1]) < x :
                #     g = 1

                cur = r

                # print(cur,j,x,val,lvl,g)

                if cur >= j :
                    if lvl-g <= val :
                        while more[j,x] :
                            i = more[j,x].pop()
                            ans[i] = "YES"
                    break
                if cur == n :
                    break

    for line in ans :
        print(line)

###############################################################################
###############################################################################
###############################################################################

# for t in range(int(input())):
    # case(t+1)
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 generate a challenging test input that exposes the bug in the given buggy code, let's consider the following steps:

1. **Understand the Bug**: The buggy code seems to have issues with the way it handles the `more` dictionary and the binary search logic. Specifically, the logic for determining if Monocarp will fight the monster might not be correct for large values of `k`.

2. **Edge Cases**:
   - **Monsters with the same level as Monocarp's initial level**: This can cause issues if the level-up logic is not handled correctly.
   - **Monsters with levels just above Monocarp's current level**: This can test the logic for fighting and fleeing.
   - **Large values of `k`**: This can test the binary search and cumulative sum logic.

3. **Small-Scale Test Input**:
   - We will create a small number of monsters and queries to ensure the input is manageable but still challenging.
   - We will include a mix of monsters with levels close to Monocarp's initial level and some higher levels.
   - We will include queries with both small and large values of `k`.

Here is the small-scale test input:

```plaintext
5 10
1 2 3 4 5
1 1
2 1
3 1
4 1
5 1
1 5
2 5
3 5
4 5
5 5
```

This input includes:
- 5 monsters with levels 1, 2, 3, 4, and 5.
- 10 queries with a mix of small and large values of `k` (1 and 5).

This should help expose any issues in the buggy code, especially with the handling of large `k` values and the logic for determining if Monocarp will fight a monster.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

Buggy Code:
import sys
import math
import bisect
input = sys.stdin.readline

n, q = list(map(int, input().split()))
a = list(map(int, input().split()))
sqrtn = int(math.sqrt(n))

dp = [0 for _ in range(sqrtn)]
for k in range(1, sqrtn + 1):
    cur_level = 1
    cur_cnt = 0
    for i in range(n):
        if a[i] >= cur_level:
            dp[k - 1] |= (1 << i)
            cur_cnt += 1
            if cur_cnt == k:
                cur_level += 1
                cur_cnt = 0

dp2 = [[-1] for _ in range(sqrtn)]
for level in range(1, sqrtn):
    for i in range(n):
        if a[i] >= level:
            dp2[level - 1].append(i)

for _ in range(q):
    i, x = list(map(int, input().split()))
    if x <= sqrtn:
        print("YES" if dp[x - 1] & (1 << (i - 1)) else "NO")
    else:
        if a[i - 1] >= 1 + (i - 1) // x:
            print("YES")
            continue
        target_level = a[i - 1]
        cur_level = 1
        cur_idx = 0
        while cur_idx < n and cur_idx < i:
            pre = bisect.bisect_left(dp2[cur_level - 1], cur_idx) - 1
            nxt_idx = dp2[cur_level - 1][pre + x] + 1 if pre + x < len(dp2[cur_level - 1]) else n
            if nxt_idx > i - 1:
                print("YES" if target_level >= cur_level else "NO")
                break
            cur_idx = nxt_idx
            cur_level += 1
            if cur_level > target_level:
                print("NO")
                break

**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 code, we need to consider scenarios that might cause the code to fail. The code has a specific approach for handling small values of `k` (up to `sqrt(n)`) and a different approach for larger values of `k`. We need to focus on edge cases and scenarios that might cause the code to misbehave.

### Key Points to Consider:
1. **Small `k` Values**: The code uses a bitmask to track which monsters are fought. We need to ensure that the bitmask is correctly updated and that the logic for determining the level and fight count is correct.
2. **Large `k` Values**: The code uses a binary search to find the next fight index. We need to ensure that the binary search logic and the level update logic are correct.
3. **Monsters with the Same Level**: This can cause issues in the binary search logic if not handled correctly.
4. **Monsters with Levels Close to the Initial Level**: This can cause issues in the level update logic.
5. **Monsters with Levels Much Higher Than the Initial Level**: This can cause issues in the fight count logic.

### Challenging Test Input:
Let's create a test input that includes:
- Monsters with the same level.
- Monsters with levels close to the initial level.
- Monsters with levels much higher than the initial level.
- Queries that test both small and large values of `k`.

```plaintext
7 10
1 1 2 1 1 3 3
1 1
2 1
3 1
4 1
5 1
6 1
7 1
1 5
6 5
7 5
```

### Explanation:
- **Monsters**: The levels are `[1, 1, 2, 1, 1, 3, 3]`.
  - Monocarp starts at level 1.
  - The first five monsters have levels 1 or 2, which are close to Monocarp's initial level.
  - The last two monsters have levels 3, which are higher than Monocarp's initial level.
- **Queries**:
  - Queries for `k = 1` to ensure the bitmask logic is correct.
  - Queries for `k = 5` to ensure the binary search logic is correct.

This test input should help expose any bugs in the code, especially in the handling of small and large `k` values, and in the logic for updating the level and fight count.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

Buggy Code:
//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx2")

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;

int bubbleGum = 0;
#ifdef RODYA
#define FAST_IO 0;
#include "Templates/Debug template.cpp"
#else
#define FAST_IO ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
#define dbg(...) ++bubbleGum
#define dbgArr(x) ++bubbleGum
#endif

// #define int ll
#define vec vector
#define eb emplace_back
#define pb push_back
#define sz(v) (int)(v.size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define mp(a, b) make_pair(a, b)
#define endl "\n"

// BE CAREFUL WITH THESE STUFF
#define YESr { cout << "YES" << endl; return; }
#define NOr { cout << "NO" << endl; return; };
#define Yesr { cout << "Yes" << endl; return; };
#define Nor { cout << "No" << endl; return; };
#define yesr { cout << "yes" << endl; return; };
#define nor { cout << "no" << endl; return; };

template<class T> istream& operator>>(istream& i, vec<T>& v) { for (auto &it: v) i >> it; return i; }

using pii = pair<int, int>;

const int N = 2e5 + 10, INF = 1e9 + 10, MEM = 4e6 + 10;

struct Node {
    int le = 0, ri = 0, sum = 0;
    Node() {}
    Node(int le, int ri, int sum) : le(le), ri(ri), sum(sum) {}
};

int nodeInd = 1;

int newNode() {
    int res = nodeInd++;
    assert(res < MEM);
    return res;
}

Node nodes[MEM];
int versions[N];
int n, q;
int a[N];
vec<pii> inde[N];
int answer[N];

struct SegmentTree {
    SegmentTree() {}
    void build(int v, int tl, int tr) {
        nodes[v].sum = 0;
        if (tl == tr) return;
        int le = newNode(), ri = newNode();
        int mid = (tl + tr) / 2;
        nodes[v].le = le, nodes[v].ri = ri;
        build(nodes[v].le, tl, mid);
        build(nodes[v].ri, mid + 1, tr);
    }
    int update(int v, int tl, int tr, int ind, int dx) {
        int cop = newNode();
        nodes[cop] = nodes[v];

        nodes[cop].sum += dx;
        if (tl == tr) return cop;
        int mid = (tl + tr) / 2;
        if (ind <= mid) {
            nodes[cop].le = update(nodes[cop].le, tl, mid, ind, dx);
        } else {
            nodes[cop].ri = update(nodes[cop].ri, mid + 1, tr, ind, dx);
        }
        return cop;
    }
    int query(int v1, int v2, int tl, int tr, int l, int r) {
        if (r < tl || tr < l) return 0;
        if (l <= tl && tr <= r) {
            return nodes[v2].sum - nodes[v1].sum;
        }

        int mid = (tl + tr) / 2;
        int one = query(nodes[v1].le, nodes[v2].le, tl, mid, l, r);
        int two = query(nodes[v1].ri, nodes[v2].ri, mid + 1, tr, l, r);
        return one + two;
    }
} tree;

void input() {
    cin >> n >> q;
    for (int i = 1; i <= n; ++i)
        cin >> a[i];
}

void prepare() {
    versions[0] = newNode();
    for (int i = 1; i <= n; ++i) {
        versions[i] = tree.update(versions[i - 1], 1, N - 1, a[i], 1);
    }
}

void solve() {
    for (int i = 1; i <= q; ++i) {
        int ind, k; cin >> ind >> k;
        inde[k].eb(ind, i);
    }

    for (int k = 1; k <= n; ++k) {
        if (inde[k].empty()) continue;

        int cur = 1, pow = 1;
        vec<pii> segms; vec<int> pows;
        dbg(k, "--------------------------");
        while (true) {
            int l = cur - 1, r = n + 1;
            while (r > l + 1) {
                int mid = (l + r) / 2;
                int cnt = tree.query(versions[cur - 1], versions[mid], 1, N - 1, pow, N - 1);
                dbg(cur, mid, pow, cnt);
                if (cnt < k) l = mid;
                else r = mid;
            }

            // int cnt = tree.query(versions[cur - 1], versions[n], 1, N - 1, pow, N - 1);
            // if (!cnt) {
            //     segms.eb(cur, n);
            //     pows.eb(pow);
            //     break;
            // }

            dbg(cur, r);
            //bs.eb(cur, r, pow);
            segms.eb(cur, r);
            pows.eb(pow);

            cur = r + 1;
            pow += 1;

            if (cur > n) break;
        }
        dbg(segms);
        dbg(pows);

        for (auto& [ind, queryInd] : inde[k]) {
            int i = lower_bound(all(segms), mp(ind, INF)) - segms.begin() - 1;
            dbg(ind, queryInd, i, segms[i], pows[i], a[ind]);
            bool val = a[ind] >= pows[i];
            answer[queryInd] = val;
        }
    }

    for (int i = 1; i <= q; ++i)
        cout << (answer[i] ? "YES" : "NO") << "\n";
}

void test() {

    input();
    prepare();
    solve();

}

signed main() {
#ifdef RODYA
    setLocalThings();
#endif
    FAST_IO

    int tests = 1; // cin >> tests;
    for (int testCase = 1; testCase <= tests; ++testCase) {
        dbg(testCase);
        test();
    }
}

/*

 */

**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 code, we need to consider edge cases and scenarios that might cause the code to fail. Here are some key points to consider:

1. **Monsters with the same level**: This can test how the code handles multiple monsters of the same level.
2. **Monsters with increasing levels**: This can test the level-up mechanism.
3. **Monsters with decreasing levels**: This can test how the code handles levels that decrease.
4. **Queries with different values of \( k \)**: This can test the code's ability to handle different level-up thresholds.
5. **Queries at the boundaries**: This can test edge cases where the monster is the first or last in the list.

Given these considerations, let's create a small-scale test input:

```plaintext
5 10
1 2 3 2 1
1 1
2 1
3 1
4 1
5 1
1 2
2 2
3 2
4 2
5 2
```

This test input includes:
- 5 monsters with levels [1, 2, 3, 2, 1].
- 10 queries:
  - The first 5 queries test the behavior with \( k = 1 \).
  - The next 5 queries test the behavior with \( k = 2 \).

This input will help expose any issues in the level-up mechanism and the handling of different \( k \) values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

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

using namespace std;

int N = 200000;

int n,q;
int a[200005];
vector<int> f[200005];///f(i,k) totusi
vector<int> pozi[200005];

int aib[200005];

void update(int pos, int val)
{
    for (int i = pos; i <= n; i += (i & -i))
        aib[i] += val;
}

int query(int pos)
{
    int rr = 0;
    for (int i = pos; i > 0; i -= (i & -i))
        rr += aib[i];
    return rr;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    cin >> n >> q;
    for (int i = 1; i <= n; i++)
        cin >> a[i],a[i]--,pozi[a[i]].push_back(i);
    f[0].resize(N + 3);
    for (int i = 1; i <= N; i++)
        f[i].resize(N / i + 3);
    for (int k = 1; k <= N; k++)
        f[0][k] = 0;
    for (int i = 1; i <= n; i++)
        update(i,1);
    for (int i = 1; i <= n; i++)
    {
        if (i >= 2)
        {
            for (auto it : pozi[i - 2])
                update(it,-1);
        }
        for (int k = 1; k * i <= n; k++)
        {
            if (k >= f[i].size() or k >= f[i - 1].size())
                while(true);
            int pz = f[i - 1][k];
            if (pz == n + 1)
                f[i][k] = n + 1;
            else
            {
                ///vreau al k-lea 1 dupa pz
                int ct = k + query(pz);
                int st = 0,pas = 1 << 17;
                int hm = 0;
                while (pas != 0)
                {
                    if (st + pas <= n)
                    {
                        if (hm + aib[st + pas] < ct)
                            hm += aib[st + pas], st += pas;
                    }
                    pas /= 2;
                }
                f[i][k] = st + 1;
            }
            //cout << i << ' ' << k << ' ' << f[i][k] << endl;
        }
    }
    for (int i = 1; i <= q; i++)
    {
        int p,k;
        cin >> p >> k;
        if ((a[p] + 1) * k >= p)
            cout << "YES\n";
        else
        {
            if (k >= f[a[p] + 1].size())
                while(true);
            if (f[a[p] + 1][k] >= p)
                cout << "YES\n";
            else
                cout << "NO\n";
        }
    }
    return 0;
}

///n-a venit spatarel :(

/**
vai vai, au teste bune
pai atunci sol legita: evident calculez pentru (k,i) cand ajung la puterea i pentru k -> NlogN perechi
Daca stiu (k,i) si vreau sa aflu (k,i + 1), vreau a k-a valoare >= i de la f(k,i) incolo
Pai vad cate valori >= i exista pana la f(k,i) (fie ele x) si vreau a (x+k)-a valoare >= i din sir
Daca calculez si eu dupa i crescator asta e doar aib yayy
**/

**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 where the code might fail. Here are the steps to think through this:

1. **Understand the Bug**: The provided code seems to have issues with how it handles the level up mechanism and the queries. Specifically, it might not correctly account for the number of fights required to level up, especially when the levels of monsters are not in a straightforward increasing or decreasing order.

2. **Edge Cases**:
   - **Monsters with the same level**: This can cause issues if the code does not correctly handle multiple monsters at the same level.
   - **Monsters with levels that are close to each other**: This can cause issues with the binary search and AIB (Binary Indexed Tree) calculations.
   - **Monsters with levels that are much higher than the initial level**: This can cause issues with the level up mechanism.

3. **Small-Scale Test Input**: To ensure the test input is small and valid, we can create a scenario with a few monsters and a few queries that should highlight the issues in the code.

Here is a small-scale test input that should expose the bug:

```plaintext
5 10
1 2 2 1 3
1 1
2 1
3 1
4 1
5 1
1 2
2 2
3 2
4 2
5 2
```

### Explanation:
- **Monsters**: The levels of the monsters are [1, 2, 2, 1, 3].
- **Queries**:
  - For \( k = 1 \):
    - Query 1: \( i = 1 \) (YES, as Monocarp's level is 1 and the monster's level is 1)
    - Query 2: \( i = 2 \) (YES, as Monocarp's level is 1 and the monster's level is 2)
    - Query 3: \( i = 3 \) (NO, as Monocarp's level is 1 and the monster's level is 2)
    - Query 4: \( i = 4 \) (YES, as Monocarp's level is 2 and the monster's level is 1)
    - Query 5: \( i = 5 \) (NO, as Monocarp's level is 2 and the monster's level is 3)
  - For \( k = 2 \):
    - Query 1: \( i = 1 \) (YES, as Monocarp's level is 1 and the monster's level is 1)
    - Query 2: \( i = 2 \) (NO, as Monocarp's level is 1 and the monster's level is 2)
    - Query 3: \( i = 3 \) (NO, as Monocarp's level is 1 and the monster's level is 2)
    - Query 4: \( i = 4 \) (YES, as Monocarp's level is 2 and the monster's level is 1)
    - Query 5: \( i = 5 \) (NO, as Monocarp's level is 2 and the monster's level is 3)

This test input should expose issues in the code's handling of the level up mechanism and the binary search/AIB calculations.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

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

public class Main {
    static PrintWriter pw = new PrintWriter(System.out);
    static Scanner sc = new Scanner(System.in);

    public static void solve() throws IOException {
        int n = sc.nextInt(), q = sc.nextInt();
        int a [] = sc.nextIntArray(n);
        int prefix [][] = new int [501][n];
        for(int i = 0; i < n ; i++){
            for(int j = 1 ; j <= 500 ; j++) {
                if (a[i] == j) {
                    prefix[j][i]++;
                }
            }
        }
        for(int i = 0; i < 500 ; i++){
            for(int j = 1 ; j < n ; j++) {
                prefix[i][j] += prefix[i][j - 1];
            }
        }
        for(int i = 0; i < n ; i++){
            for(int j = 1 ; j <= 500 ; j++) {
                prefix[j][i] += prefix[j - 1][i];
            }
        }
        while(q -- > 0){
            int idx = sc.nextInt() - 1, k = sc.nextInt();
//            if(k <= 500) {
//                pw.println(b[idx][k] == 1 ? "YES" : "NO");
//            }
            {
                int curr = 0, pointer = 0;
                while(curr <= 500){
                    int low = pointer, high = n - 1;
                    while(low <= high){
                        int mid = (low + high) / 2;
                        if(mid - pointer + 1 - (prefix[curr][mid] - (pointer == 0 ? 0 : prefix[curr][pointer - 1])) >= k)
                            high = mid - 1;
                        else
                            low = mid + 1;
                    }
                    curr++;
                    if(idx <= low){
                        if(a[idx] >= curr)
                            pw.println("YES");
                        else
                            pw.println("NO");
                        break;
                    }
                    pointer = low + 1;
                }
            }
        }
    }

    public static void main(String[] args) throws IOException {
//        pw.println(200000 +" " + 1);
//        for(int i = 0; i < 200000 ; i++)
//            pw.print(1+" ");
//        pw.println();
//        pw.println(200000+" " + 600);
//        pw.flush();
        int t = 1;
        while(t -- > 0)
            solve();
        pw.flush();
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

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

        public Scanner(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(s));
        }

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

        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();
        }
    }
}

**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 are likely to cause issues. Here are the steps to think through the process:

1. **Understand the Bug**: The buggy code seems to have issues with handling large values of `k` and the cumulative level-up logic. It uses a prefix sum array but may not correctly handle the level-up conditions for large `k`.

2. **Edge Cases**:
   - **Small `n` and `q`**: Ensure the test case is small enough to be manageable.
   - **Monsters with levels just above the current level**: This can test the level-up logic.
   - **Queries with large `k`**: This can expose issues with the cumulative level-up logic.
   - **Queries with `k` equal to `n`**: This can test the edge case where the level-up happens exactly at the end.

3. **Construct the Test Case**:
   - **Number of Monsters (`n`)**: 5
   - **Number of Queries (`q`)**: 5
   - **Monsters' Levels**: [1, 2, 2, 3, 3]
   - **Queries**:
     - (1, 1): Monocarp starts at level 1, and the first monster has level 1, so he fights.
     - (2, 2): Monocarp starts at level 1, and the second monster has level 2, so he fights. After 2 fights, his level becomes 2, and the second monster flees.
     - (3, 3): Monocarp starts at level 1, and the third monster has level 2, so he fights. After 3 fights, his level becomes 2, and the third monster flees.
     - (4, 4): Monocarp starts at level 1, and the fourth monster has level 3, so he fights. After 4 fights, his level becomes 2, and the fourth monster flees.
     - (5, 5): Monocarp starts at level 1, and the fifth monster has level 3, so he fights. After 5 fights, his level becomes 2, and the fifth monster flees.

Here is the test input:

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

This test input should expose the bug in the cumulative level-up logic and the handling of large `k` values.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Level Up
time_limit_ms: 4000
memory_limit_mb: 512
Description: Monocarp is playing a computer game. He starts the game being level $1$. He is about to fight $n$ monsters, in order from $1$ to $n$. The level of the $i$-th monster is $a_i$.

For each monster in the given order, Monocarp's encounter goes as follows:

  * if Monocarp's level is strictly higher than the monster's level, the monster flees (runs away);    * otherwise, Monocarp fights the monster.

After every $k$-th fight with a monster (fleeing monsters do not count), Monocarp's level increases by $1$. So, his level becomes $2$ after $k$ monsters he fights, $3$ after $2k$ monsters, $4$ after $3k$ monsters, and so on.

You need to process $q$ queries of the following form:

  * $i~x$: will Monocarp fight the $i$-th monster (or will this monster flee) if the parameter $k$ is equal to $x$?

Input: The first line contains two integers $n$ and $q$ ($1 \le n, q \le 2 \cdot 10^5$) — the number of monsters and the number of queries.

The second line contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \le a_i \le 2 \cdot 10^5$) — the levels of the monsters.

In the $j$-th of the following $q$ lines, two integers $i$ and $x$ ($1 \le i, x \le n$) — the index of the monster and the number of fights required for a level up in the $j$-th query.

Output: For each query, output "YES", if Monocarp will fight the $i$-th monster in this query, and "NO", if the $i$-th monster flees.

Examples:
input:
4 16
2 1 2 1
1 1
2 1
3 1
4 1
1 2
2 2
3 2
4 2
1 3
2 3
3 3
4 3
1 4
2 4
3 4
4 4
output:
YES
NO
YES
NO
YES
YES
YES
NO
YES
YES
YES
NO
YES
YES
YES
YES

input:
7 15
1 1 2 1 1 1 1
5 3
2 2
2 2
1 6
5 1
5 5
7 7
3 5
7 4
4 3
2 5
1 2
5 6
4 1
6 1
output:
NO
YES
YES
YES
NO
YES
YES
YES
NO
NO
YES
YES
YES
NO
NO

Note:

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

public class Main {
    static PrintWriter pw = new PrintWriter(System.out);
    static Scanner sc = new Scanner(System.in);

    public static void solve() throws IOException {
        int n = sc.nextInt(), q = sc.nextInt();
        int a [] = sc.nextIntArray(n);
        boolean b[][] = new boolean [n][701];
        for(int j = 1 ; j <= 700 ; j++) {
            int x = j, te = x, curr = 1;
            for (int i = 0; i < n; i++) {
                if(a[i] >= curr){
                    b[i][j] = true;
                    te--;
                }
                if(te == 0){
                    curr++;
                    te = x;
                }
            }
        }
        int prefix [][] = new int [288][n];
        for(int i = 0; i < n ; i++){
            for(int j = 1 ; j <= 287 ; j++) {
                if (a[i] == j) {
                    prefix[j][i]++;
                }
            }
        }
        for(int i = 0; i <= 287 ; i++){
            for(int j = 1 ; j < n ; j++) {
                prefix[i][j] += prefix[i][j - 1];
            }
        }
        for(int i = 0; i < n ; i++){
            for(int j = 1 ; j <= 287 ; j++) {
                prefix[j][i] += prefix[j - 1][i];
            }
        }
        while(q -- > 0){
            int idx = sc.nextInt() - 1, k = sc.nextInt();
            if(k <= 700) {
                pw.println(b[idx][k] ? "YES" : "NO");
            }
            else{
                int curr = 0, pointer = 0;
                while(curr <= 287){
                    int low = pointer, high = n - 1;
                    while(low <= high){
                        int mid = (low + high) / 2;
                        if(mid - pointer + 1 - (prefix[curr][mid] - (pointer == 0 ? 0 : prefix[curr][pointer - 1])) >= k)
                            high = mid - 1;
                        else
                            low = mid + 1;
                    }
                    curr++;
                    if(idx <= low){
                        if(a[idx] >= curr)
                            pw.println("YES");
                        else
                            pw.println("NO");
                        break;
                    }
                    pointer = low + 1;
                }
            }
        }
    }

    public static void main(String[] args) throws IOException {
        int t = 1;
        while(t -- > 0)
            solve();
        pw.flush();
    }

    static class Scanner {
        StringTokenizer st;
        BufferedReader br;

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

        public Scanner(String s) throws FileNotFoundException {
            br = new BufferedReader(new FileReader(s));
        }

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

        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();
        }
    }
}

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — the number of problems in the olympiad.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the scores of the problems.

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

It is guaranteed that the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define INF (int)1e18
#define fi first
#define se second
#define pii pair<int,int>
const int MOD = 1e9 + 7;
mt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());

#ifndef ONLINE_JUDGE
#include "debugtemp.cpp"
#else
#define debug(...)
#define debugArr(...)
#endif

void Solve()
{
    int n; cin >> n;
    vector <int> a(n) , b(n);
    for(auto &x : a) cin >> x;
    for(auto &x : b) cin >> x , x--;
    vector <array<int, 2>> dp(n , {0 , 0});
    vector <bool> vis(n , false);
    multiset <int> m;
    for(int i = 0 ; i < n ; i++){
        m.insert(i);
    }
    auto dfs = [&](auto && self , int u) -> void{
        if(vis[u]){
            return;
        }
        vis[u] = true;
        int v = -1;
        for(int v = b[u] ; v >= 0  ; v--){
            if(!vis[v]){
                self(self , v);
                dp[u][0] += max(dp[v][0] , dp[v][1]);
                break;
            }
        }
        if(v != -1){
            vis[v] = false;
        }
        dp[u][1] += a[u];
        for(int v = u - 1 ; v >= 0 ; v--){
            if(!vis[v]){
                self(self , v);
                dp[u][1] += max(dp[v][0] , dp[v][1]);
                break;
            }
        }
        // m.erase(u);
        // if(m.size() > 0){
        //     auto it = m.upper_bound(b[u]);
        //     if(it == m.begin()){
        //         dp[u][0] = 0;
        //     }
        //     else{
        //         it--;
        //         self(self , *(it));
        //         dp[u][0] = max(dp[*(it)][0] , dp[*(it)][1]);
        //     }
        //     auto it1 = m.upper_bound(u);
        //     if(it1 == m.begin()){
        //         dp[u][1] += a[u];
        //     }
        //     else{
        //         it1--;
        //         self(self , *(it1));
        //         dp[u][1] += max(dp[*(it1)][0] , dp[*(it1)][1]) + a[u];
        //     }
        // }
        // else{
        //     dp[u][1] += a[u];
        // }

    };

    dfs(dfs , 0);
    debug(dp);
    cout << max(dp[0][0] , a[0]) << "\n";
}
int32_t main()
{
    auto begin = std::chrono::high_resolution_clock::now();
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    #ifndef ONLINE_JUDGE
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout);
    #endif
    int t = 1;
    cin >> t;
    for (int i = 1; i <= t; i++)
    {
        // cout << "Case #" << 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
4
10 100 1000 10000
2 3 4 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — the number of problems in the olympiad.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the scores of the problems.

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

It is guaranteed that the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

Buggy Code:
#include<bits/stdc++.h>
using namespace std;

#define int long long
const int N = 4e5 + 10;

int a[N],b[N];

void cgy() {
	int n;
	cin>>n;
	for(int i = 1;i <= n;i++) cin>>a[i];
	for(int i = 1;i <= n;i++) cin>>b[i];
	vector<int>sum(n + 1,0);
	for(int i = 1;i <= n;i++){
		sum[i] = sum[i - 1] + a[i];
	}
	//cout<<"n = "<<n<<'\n';
	int pre = 0;
	int ans = a[1];
	if(b[1] == 1){
		cout<<a[1]<<'\n';
		return;
	}
	int now_pos = 1;
	int pre_pos = 1,mx_pos= now_pos;
	int summm = sum[1];
	int mx = 0;
	while(1){
		//cout<<"pre = "<<pre<<" now_pos = "<<now_pos<<" pre + sum = "<<pre + sum[now_pos]<<'\n';
		now_pos = b[now_pos];
		ans = max(ans,pre + summm);
		pre -= a[mx_pos];
		ans = max(ans,pre + sum[now_pos]);
		mx = 0,mx_pos = 0;

		for(int i = now_pos;i > pre_pos;i--){
			if(b[i] > mx){
				mx = b[i];
				mx_pos = i;
			}
			else if(b[i] == mx){
				if(a[mx_pos] > a[i]){
					mx_pos = i;
				}
			}
		}
		//cout<<"mx_pos = "<<mx_pos<<" bp = "<<b[mx_pos]<<" now_pos = "<<now_pos<<" pre = "<<pre<<'\n';

		if(mx_pos == 0 || b[mx_pos] <= now_pos){
			//ans = max({ans,pre + sum[now_pos]});
			break;
		}

		summm = sum[now_pos];

		pre_pos = mx_pos;

		//cout<<"now_pos = "<<now_pos<<" summm = "<<summm<<" pre = "<<pre<<'\n';
		now_pos = mx_pos;
	}
	cout<<ans<<'\n';
}
signed main() {
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int T = 1;
	cin >> T;
	while (T--) cgy();
}

**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
100 200 300 400 500
5 4 3 2 1
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Skipping
time_limit_ms: 2000
memory_limit_mb: 256
Description: It is already the year $3024$, ideas for problems have long run out, and the olympiad now takes place in a modified individual format. The olympiad consists of $n$ problems, numbered from $1$ to $n$. The $i$-th problem has its own score $a_i$ and a certain parameter $b_i$ ($1 \le b_i \le n$).

Initially, the testing system gives the participant the first problem. When the participant is given the $i$-th problem, they have two options:

  * They can submit the problem and receive $a_i$ points;    * They can skip the problem, in which case they will never be able to submit it.

Then, the testing system selects the next problem for the participant from problems with indices $j$, such that:

  * If he submitted the $i$-th problem, it looks at problems with indices $j < i$;    * If he skipped the $i$-th problem, it looks at problems with indices $j \leq b_i$.

Among these problems, it selects the problem with the maximum index that it has not previously given to the participant (he has neither submitted nor skipped it before). If there is no such problem, then the competition for the participant ends, and their result is equal to the sum of points for all submitted problems. In particular, if the participant submits the first problem, then the competition for them ends. Note that the participant receives each problem at most once.

Prokhor has prepared thoroughly for the olympiad, and now he can submit any problem. Help him determine the maximum number of points he can achieve.

Input: Each test consists of multiple test cases. The first line contains a single integer $t$ ($1 \leq t \leq 10^5$) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer $n$ ($1 \leq n \leq 4 \cdot 10^5$) — the number of problems in the olympiad.

The second line of each test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \leq a_i \leq 10^9$) — the scores of the problems.

The third line of each test case contains $n$ integers $b_1, b_2, \ldots, b_n$ ($1 \leq b_i \leq n$) — the parameters of the problems.

It is guaranteed that the sum of $n$ over all test cases does not exceed $4 \cdot 10^5$.

Output: For each test case, output a single integer — the maximum number of points that Prokhor can achieve.

Examples:
input:
4
2
15 16
2 1
5
10 10 100 100 1000
3 4 1 1 1
3
100 49 50
3 2 2
4
100 200 300 1000
2 3 4 1
output:
16
200
100
1000

Note: In the first test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 2$. Prokhor can submit it and receive $a_2 = 16$ points. After that, the competition will end because Prokhor has already received all problems. Note that if Prokhor submits the first problem, he will receive $a_1 = 15$ points, but the competition will end immediately.

In the second test case, Prokhor can skip the first problem; then he will receive the problem with index $b_1 = 3$. Prokhor can submit it and receive $a_3 = 100$ points. After that, Prokhor will receive the second problem, which he can skip to receive the problem with index $b_2 = 4$. Prokhor can submit the fourth problem and receive another $a_4 = 100$ points. After that, the competition ends because Prokhor has already received all problems with indices not exceeding $4$. Thus, Prokhor will receive a total of $200$ points.

In the third test case, Prokhor can submit the first problem and receive $100$ points, after which the competition will end immediately.

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
vector<int> a, b, dp;
int n;

int d(int i){
    if(i == 0){
        return 0;
    }
    if(dp[i] == -1){
        int sc = a[i];
        dp[i] = -2;
        if(i > 1){
            sc += d(i - 1);
        }
        dp[i] = max(sc, d(b[i]));
        return dp[i];
    }else{
        if(dp[i] == -2){
            return 0;
        }
        return dp[i];
    }
}

main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    int t; cin >> t;
    while(t--){
        cin >> n;
        dp.clear();
        a.resize(n+1);
        b.resize(n+1);
        dp.resize(n+1, -1);
        for(int i = 1; i <= n; i++){
            cin >> a[i];
        }
        for(int i = 1; i <= n; i++){
            cin >> b[i];
        }
        cout << d(1) << '\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
2
10 20
2 1
3
10 100 1
3 2 2
4
100 200 300 1000
2 3 4 1
```