**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:
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashSet;
import java.util.List;
import java.util.TreeSet;

public class Main {

	static class Scanner {
		Scanner(InputStream in) { this.in = in; } InputStream in;
		byte[] bb = new byte[200000 * 19]; int i, n;
		byte getc() {
			if (i == n) {
				i = n = 0;
				try { n = in.read(bb); } catch (IOException e) {}
			}
			return i < n ? bb[i++] : 0;
		}
		int nextInt() {
			byte c = 0; while (c <= ' ') c = getc();
			boolean negate = false;
			if (c == '-') {
				negate = true;
				c = getc();
			}
			int a = 0; while (c > ' ') { a = a * 10 + c - '0'; c = getc(); }
			if (negate) a = -a;
			return a;
		}

		long nextLong() {
			byte c = 0; while (c <= ' ') c = getc();
			long a = 0; while (c > ' ') { a = a * 10 + c - '0'; c = getc(); }
			return a;
		}
	}

	private static class Edge {
		int u;
		int weight;
		Edge(int u, int w) {
			this.u = u;
			this.weight = w;
		}
	}

	private static class Vertex implements Comparable<Vertex> {
		int v;
		long dist;
		Vertex(int v, long dist) {
			this.v = v;
			this.dist = dist;
		}

		@Override
		public int compareTo(Main.Vertex other) {
			if (dist < other.dist)
				return -1;
			else if (dist > other.dist)
				return 1;
			else
				return 0;
		}

		/*@Override
		public boolean equals(Object other) {
			return this.compareTo((Main.Vertex)other) == 0;
		}*/
	}

	public static void main(String[] args) throws Exception {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in), 200000 * 20);
		BufferedWriter out = new BufferedWriter(new OutputStreamWriter(System.out), 50000000);

		int T = Integer.parseInt(in.readLine());

		int[] A = new int[400400];
		int[] B = new int[400400];
		long[] dist = new long[400400];
		long[] prefixSum = new long[400400];
		List<Edge>[] G = new ArrayList[400400];

		for (int ts=1; ts<=T; ts++) {
			int N = Integer.parseInt(in.readLine());

			String[] line = in.readLine().split(" ");
			for (int i=1; i<=N; i++) {
				A[i] = Integer.parseInt(line[i-1]);
			}

			line = in.readLine().split(" ");
			for (int i=1; i<=N; i++) {
				B[i] = Integer.parseInt(line[i-1]);
			}

			for (int i=1; i<=N; i++) {
				G[i] = new ArrayList<>();
				if (B[i] > i) {
					G[i].add(new Edge(B[i], A[i]));
				}
				if (i > 1) {
					G[i].add(new Edge(i-1, 0));
				}
			}

			for (int i=1; i<=N; i++) {
				dist[i] = Long.MAX_VALUE;
			}

			prefixSum[0] = 0;
			for (int i=1; i<=N; i++) {
				prefixSum[i] = prefixSum[i-1] + A[i];
			}

			TreeSet<Vertex> pq = new TreeSet<>();
			HashSet<Integer> hs = new HashSet<>();
			pq.add(new Vertex(1, 0));
			dist[1] = 0;

			long result = 0;

			while (!pq.isEmpty()) {
				Vertex ver = pq.pollFirst();
				int v = ver.v;
				long dst = ver.dist;

				if (hs.contains(v)) {
					//throw new RuntimeException("12345");
					continue;
				}
				hs.add(v);

				//out.write(v + " " + (prefixSum[v] - dst) + "\n");
				if (prefixSum[v] - dst > result) {
					result = prefixSum[v] - dst;
				}
				//out.write(v + " " + dst + "\n");

				for (final Edge edge : G[v]) {
					long curDist = dst + edge.weight;
					if (curDist < dist[edge.u]) {
						pq.remove(new Vertex(edge.u, dist[edge.u]));

						dist[edge.u] = curDist;
						pq.add(new Vertex(edge.u, curDist));
					}
				}
			}

			out.write(result + "\n");
		}

		in.close();
		out.close();
	}
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases or scenarios that might cause the algorithm to fail. The code seems to have issues with handling the transitions between problems, especially when the participant skips problems and the next problem is chosen based on the `b_i` parameter.

Here are the steps to think about the test case:

1. **Understanding the Code Logic**:
   - The code uses a priority queue (TreeSet) to manage the problems and their distances.
   - It updates the distances and adds new vertices to the priority queue based on the edges (problems and their weights).
   - The `prefixSum` array is used to calculate the cumulative sum of scores up to each problem.

2. **Identifying Potential Issues**:
   - The code might have issues when the participant skips a problem and the next problem is chosen based on the `b_i` parameter.
   - The code might not correctly handle the case where the participant submits a problem and the next problem is chosen from problems with indices less than the current problem.

3. **Constructing the Test Case**:
   - We need a small-scale test case that maximizes the failure probability.
   - Consider a scenario where the participant skips a problem and the next problem is chosen based on `b_i`.
   - Ensure that the `b_i` values are such that the code might incorrectly choose the next problem.

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

```plaintext
3
3
10 20 30
3 2 1
2
100 200
2 1
4
10 20 30 40
4 3 2 1
```

### Explanation of the Test Case:
1. **Test Case 1**:
   - `n = 3`
   - `a = [10, 20, 30]`
   - `b = [3, 2, 1]`
   - If the participant skips the first problem, they will receive the third problem next (since `b_1 = 3`). Submitting the third problem gives 30 points. The competition ends because the second problem is not considered.

2. **Test Case 2**:
   - `n = 2`
   - `a = [100, 200]`
   - `b = [2, 1]`
   - If the participant skips the first problem, they will receive the second problem next (since `b_1 = 2`). Submitting the second problem gives 200 points. The competition ends.

3. **Test Case 3**:
   - `n = 4`
   - `a = [10, 20, 30, 40]`
   - `b = [4, 3, 2, 1]`
   - If the participant skips the first problem, they will receive the fourth problem next (since `b_1 = 4`). Submitting the fourth problem gives 40 points. The competition ends because the other problems are not considered.

This test input should help expose any issues with the logic in the code, particularly with how the next problem is chosen when the participant skips a problem.


**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;
#ifdef LOCAL
#include "algo/debug.h"
#else
#define debug(...) 42
#endif
typedef long long ll;
typedef unsigned long long ull;
const int N = 1e5 + 10, INF = 1e9;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    cout.tie(nullptr);
    int tt; cin >> tt;
    while (tt--) {
        int n; cin >> n;
        vector<ll> a(n), b(n);
        for (int i = 0; i < n; i++) cin >> a[i];
        for (int i = 0; i < n; i++) {
            cin >> b[i]; b[i]--;
        }
        priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<>> q;
        q.push({0LL, 0});
        vector<ll> dp(n, INF);
        while (!q.empty()) {
            auto [d, u] = q.top(); q.pop();
            if (dp[u] != INF) continue;
            dp[u] = d;
            if (b[u] > u) {
                q.push({d + a[u], b[u]});
            }
            if (u) {
                q.push({d, u - 1});
            }
        }
        ll sum = 0, ans = 0;
        for (int i = 0; i < n; i++) {
            sum += a[i];
            ans = max(ans, sum - dp[i]);
        }
        cout << ans << '\n';
    }
    return 0;
}

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

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

Think step by step.

Response:
```plaintext
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
```


**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:
import sys
import math
import copy
import itertools
import bisect
from collections import deque
from heapq import heappop, heappush, heapify

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

def ilst():return list(map(int,input().split()))

def flst():return list(map(float,input().split()))

def islst():return list(map(str,input().split()))

def inum():return map(int,input().split())

class IntKeyDict(dict):
    from random import randrange
    rand = randrange(1 << 62)
    def __setitem__(self, k, v): super().__setitem__(k^self.rand, v)
    def __getitem__(self, k): return super().__getitem__(k^self.rand)
    def __contains__(self, k): return super().__contains__(k^self.rand)
    def __repr__(self): return str({k: v for k, v in self.items()})
    def get(self, k, default=None): return super().get(k^self.rand, default)
    def keys(self): return [k^self.rand for k in super().keys()]
    def items(self): return [(k^self.rand, v) for k, v in super().items()]

def freq(l):
    d = IntKeyDict()
    for i in l:
        d[i] = d.get(i,0)+1
    return d

t = int(input())
for _ in range(t):
    n = int(input())
    a,b = ilst(),ilst()

    l = [[] for i in range(n)]

    for i in range(n):
        l[i].append([b[i]-1,a[i]])
        if i:
            l[i].append([i-1,0])
    vis = [0]*n
    cur = 0
    h = []
    dist = [float('inf') for i in range(n)]
    dist[0] = 0

    heappush(h,[0,0,-1])
    while len(h):
        wt,cur,par = heappop(h)

        if not vis[cur]:
            vis[cur] = 1
            # if cur != 0:
            #     dist[cur] = dist[par] + wt
            for ch,wt in l[cur]:
                # print(ch,wt)
                if dist[ch] > dist[cur] + wt:
                    dist[ch] = dist[cur] + wt
                    heappush(h,[dist[cur]+wt,ch,cur])
    # print(dist,h)
    # def dijkstra():
    # while len(h):
    #     cur = heappop(h)[1]
    #     if vis[cur]:
    #         continue
    #     vis[cur] = 1
    #     for child in l[cur]:
    #         ch,cwt = child
    #         if cwt+dist[cur] < dist[ch]:
    #             prev[ch] = cur
    #             dist[ch] = cwt+dist[cur]
    #             heappush(h,[cwt+dist[cur],ch])
    pref = [0]*n
    pref[0] = a[0]

    for i in range(1,n):
        pref[i] = pref[i-1] + a[i]

    ans = 0
    for i in range(n):
        if dist[i] != float('inf'):
            ans = max(ans,pref[i]-dist[i])

    print(ans)

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

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

Think step by step.

Response:
```plaintext
3
3
100 200 300
2 3 1
4
10 20 30 40
1 2 3 4
5
1000 2000 3000 4000 5000
1 2 3 4 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

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

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL);
#define ll long long
#define ld long double
#define forr(i,a,b) for (int i=a;i<b;i++)
#define ford(i,a,b) for (int i=a;i>b;i--)
#define fi first
#define se second
#define file "A"

const ll N=1e6+5;
const ll MOD=1e9+7;
const ll INF=1e9;

ll n,a[N],k[N],dp[N];

int main(){
  fastio

  #ifndef ONLINE_JUDGE
    freopen(file".inp","r",stdin);
    freopen(file".out","w",stdout);
  #endif

  cin>>n;

  dp[1]=2;
  forr(i,2,100000) {
    dp[i]=dp[i-1]%MOD*2;
  }

  forr(i,1,n+1) {
    cin>>a[i];
  }
  forr(i,1,n+1) {
    cin>>k[i];
    cout<<dp[min(a[i],k[i])]<<endl;
  }

  return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the buggy code, we need to focus on the following aspects:

1. **Edge Cases**: Test with values at the boundaries of the input constraints.
2. **Large Values**: Test with large values of \( n \) and \( k \) to ensure the modulo operation is handled correctly.
3. **Special Values**: Test with values where the difference between \( n \) and \( k \) is significant.

Given the constraints:
- \( 1 \le t \le 10^5 \)
- \( 2 \le n_i \le 10^5 \)
- \( 1 \le k_i < n_i \)

We will create a small-scale test input that includes:
- A small number of test cases (\( t = 5 \)).
- Values of \( n \) and \( k \) that are both small and large.
- Values of \( n \) and \( k \) where \( k \) is close to \( n \).

Here is the test input:

```plaintext
5
2 5 100 100000 100000
1 2 50 50000 99999
```

### Explanation:
1. **Test Case 1**: \( n = 2 \), \( k = 1 \)
   - This is a small value test case to ensure basic functionality.

2. **Test Case 2**: \( n = 5 \), \( k = 2 \)
   - This is another small value test case to ensure the formula is applied correctly.

3. **Test Case 3**: \( n = 100 \), \( k = 50 \)
   - This tests a larger value where \( k \) is half of \( n \).

4. **Test Case 4**: \( n = 100000 \), \( k = 50000 \)
   - This tests a very large value where \( k \) is half of \( n \).

5. **Test Case 5**: \( n = 100000 \), \( k = 99999 \)
   - This tests a very large value where \( k \) is very close to \( n \).

These test cases should help expose any issues in the buggy code, especially with the incorrect formula and the modulo operation.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
#define ll long long int
#include<bits/stdc++.h>
using namespace std;
ll mod = 1e9+7;
ll expo(ll x, ll y)
{
    ll res = 1;
    x %= mod;
    while (y)
    {
        if (y % 2)
            res = (res * x % mod) % mod;
        x = (x * x) % mod;
        y /= 2;
    }
    return res;
}
int main(){
ll t;
cin>>t;
ll arr[2][7];
for(int i=0;i<2;i++){
    for(int j=0;j<7;j++){
        cin>>arr[i][j];
    }
}
for(int i=0;i<7;i++){
    cout<<expo(2,arr[1][i])<<endl;
}
return 0;
}

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

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

Think step by step.

Response:
```plaintext
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
```

### Explanation:
1. **Test Case Selection**:
   - The test input is designed to cover a variety of scenarios, including small and large values of \( n \) and \( k \).
   - The pairs \((2, 1)\), \((5, 2)\), and \((5, 3)\) are small and easy to verify manually.
   - The pairs \((100000, 1)\), \((100000, 33333)\), \((100000, 66666)\), and \((100000, 99999)\) are large and will help in testing the limits of the algorithm.

2. **Edge Cases**:
   - **Small Values**: \((2, 1)\), \((5, 2)\), and \((5, 3)\) are small and can be easily computed to check the correctness of the algorithm.
   - **Large Values**: \((100000, 1)\), \((100000, 33333)\), \((100000, 66666)\), and \((100000, 99999)\) are large and will test the algorithm's performance and correctness under high limits.

3. **Modulo Operation**:
   - The results are printed modulo \(10^9 + 7\), which is a common requirement in competitive programming to prevent integer overflow and ensure the results fit within the limits of standard integer types.

4. **Expected Output**:
   - The expected output for the given test input is:
     ```
     2
     4
     8
     2
     326186014
     984426998
     303861760
     ```

This test input should effectively expose the bug in the buggy code by providing a mix of small and large values, and it is complete and valid according to the problem's constraints.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define ll long long
const int N = 1e5+10;
main()
{
	int t;
	cin >> t;
	int a1[t],a2[t];
	for(int i=0;i<t;i++)
		cin >> a1[i];
	for(int i=0;i<t;i++)
		cin >> a2[i];
	ll mod=1000000007;
	ll k0=1,k1=2,k2=4;
	for(int i=0;i<t;i++){
		ll prod=1;
		for(int j=0;j<a2[i];j++)
			prod=(prod%mod)*(2%mod);
		cout << prod << endl;
	}
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided code, we need to focus on the key differences between the correct and incorrect binomial coefficient formulas. The incorrect formula in the code is:

```cpp
C[n][k] = C[n][k - 1] + C[n - 1][k - 1];
```

While the correct formula should be:

```cpp
C[n][k] = C[n - 1][k] + C[n - 1][k - 1];
```

The provided code calculates the binomial coefficients using the incorrect formula, which results in different values compared to the correct formula. The code also uses a simplified approach that only calculates powers of 2, which is incorrect for binomial coefficients.

To expose this bug, we should use input pairs that highlight the difference between the correct and incorrect formulas. Here are the steps to create a small-scale, valid test input:

1. **Choose a small number of test cases** (e.g., 5) to keep the input manageable.
2. **Select pairs (n, k) that are likely to produce different results** using the incorrect formula compared to the correct formula.
3. **Include edge cases** such as small values of n and k, and values close to the upper limit of n (100,000).

Let's create the test input:

```plaintext
5
2 3 4 5 100000
1 2 2 3 50000
```

Explanation:
- **Test Case 1**: (2, 1) - Small values to start with.
- **Test Case 2**: (3, 2) - Another small pair to see the difference.
- **Test Case 3**: (4, 2) - A slightly larger pair.
- **Test Case 4**: (5, 3) - A pair with different n and k values.
- **Test Case 5**: (100000, 50000) - A large pair to ensure the code handles large values correctly.

This test input should expose the bug in the provided code by producing different results compared to the correct binomial coefficients.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
t = int(input())
linen = input().split()
n = [int(x) for x in linen]
linek = input().split()
k = [int(x) for x in linek]

for i in k:
    x = 2
    for j in range(1, i):
        x *= 2
        x %= (1000000000 + 7)
    print(x)

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that maximize the failure probability. The buggy code incorrectly calculates the binomial coefficients using a simple power of 2, which is not the correct approach according to the problem's description.

Here are the steps to generate a suitable test input:

1. **Small-scale Test Input**: We need a small number of pairs to keep the output manageable.
2. **Edge Cases**: We should include pairs that test the limits of the input constraints, such as the smallest and largest values for \( n \) and \( k \).
3. **Diverse Values**: Include a mix of small and large values to ensure the code handles different scenarios correctly.

Let's construct a test input with the following pairs:
- A small \( n \) and \( k \) (e.g., \( n = 2 \), \( k = 1 \))
- A larger \( n \) and \( k \) (e.g., \( n = 10 \), \( k = 5 \))
- A pair with \( n \) and \( k \) close to the maximum limit (e.g., \( n = 100000 \), \( k = 50000 \))

Here is the test input:

```plaintext
3
2 10 100000
1 5 50000
```

This test input includes:
- \( t = 3 \) pairs
- \( n_1 = 2 \), \( k_1 = 1 \)
- \( n_2 = 10 \), \( k_2 = 5 \)
- \( n_3 = 100000 \), \( k_3 = 50000 \)

This input should help expose the bug in the buggy code, as it covers a range of values and edge cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
test = int(input())

n = list(map(int,input().split()))
k=list(map(int,input().split()))

t=0
while(t<test):

    i=0
    num = 1
    # print("a")
    while(i<k[t]):
        num = num * 2
        i=i+1

    print(num%((10**9)+7))

    t=t+1

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to consider the following:

1. **Understanding the Bug**: The buggy code calculates \( C[n][k] \) using the formula \( C[n][k] = 2^k \), which is incorrect according to the problem statement. The correct formula should be \( C[n][k] = C[n-1][k] + C[n-1][k-1] \).

2. **Edge Cases**:
   - **Small values of \( n \) and \( k \)**: These can help verify the basic behavior of the code.
   - **Large values of \( n \) and \( k \)**: These can test the code's performance and correctness under stress.
   - **Boundary values**: Values where \( k = 1 \) or \( k \) is close to \( n \).

3. **Modulo Operation**: Since the output is taken modulo \( 10^9 + 7 \), we need to ensure that the values are large enough to trigger the modulo operation.

Given these considerations, a small-scale test input that includes a mix of small, large, and boundary values would be effective. Here is the test input:

```plaintext
5
2 5 100 100000 100000
1 2 50 1 99999
```

### Explanation:
- **Test Case 1**: \( n = 2 \), \( k = 1 \)
  - Expected output: 2
- **Test Case 2**: \( n = 5 \), \( k = 2 \)
  - Expected output: 4
- **Test Case 3**: \( n = 100 \), \( k = 50 \)
  - Expected output: A large number modulo \( 10^9 + 7 \)
- **Test Case 4**: \( n = 100000 \), \( k = 1 \)
  - Expected output: 2
- **Test Case 5**: \( n = 100000 \), \( k = 99999 \)
  - Expected output: A large number modulo \( 10^9 + 7 \)

This test input will help in verifying the correctness of the buggy code and should expose the incorrect formula being used.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
import sys
sys.set_int_max_str_digits(1000000)
n = int(input())
l1 = list(map(int,input().split()))
l2 = list(map(int,input().split()))
for i in range(n):
    if l1[1]==l2[i]:
        print(1)
    else:
        print((2**(l2[i]))%(10**9+7))

**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 5 10
1 2 5
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
import java.util.*;

public class App {
    public static void main(String[] args){
        Scanner sc=new Scanner(System.in);
       int n=sc.nextInt();
       int[]arr=new int[n];
       for(int j=0;j<n;j++){
        int p=sc.nextInt();
       }
       for(int j=0;j<n;j++){
        arr[j]=sc.nextInt();
       }
       for(int j=0;j<n;j++){
        long l=1;
        for(int p=0;p<arr[j];p++){
            l=l%1000000007*2;
        }
        System.out.println(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:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the incorrect formula produces significantly different results from the correct formula. The buggy code incorrectly calculates the binomial coefficients using the formula `C[n][k] = C[n][k - 1] + C[n - 1][k - 1]` instead of the correct formula `C[n][k] = C[n - 1][k] + C[n - 1][k - 1]`.

### Key Points to Consider:
1. **Small-scale input**: The input should be small enough to fit the output length limit.
2. **Edge cases**: Values of `n` and `k` that are close to the limits (e.g., `n = 2` and `k = 1`, or `n = 100000` and `k = 99999`).
3. **Modulo operation**: Ensure the values are large enough to require the modulo operation.

### Chosen Test Input:
- **Number of pairs (`t`)**: 4
- **Pairs**:
  - `(n = 2, k = 1)`
  - `(n = 5, k = 3)`
  - `(n = 10, k = 5)`
  - `(n = 100000, k = 99999)`

These pairs are chosen to cover a range of values, including small and large `n` and `k`, to highlight the differences in the results due to the incorrect formula.

### Test Input:
```plaintext
4
2 5 10 100000
1 3 5 99999
```

### Explanation:
- **Pair (2, 1)**: The correct binomial coefficient is 2, but the buggy code will also produce 2. This is a simple case to start.
- **Pair (5, 3)**: The correct binomial coefficient is 10, but the buggy code will produce 8. This is a more significant difference.
- **Pair (10, 5)**: The correct binomial coefficient is 252, but the buggy code will produce a much larger number.
- **Pair (100000, 99999)**: The correct binomial coefficient is a very large number, but the buggy code will produce a number that is significantly different and will require the modulo operation.

This test input should effectively expose the bug in the buggy code by showing the differences in the results.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
import java.util.*;
import static java.lang.System.out;

public class contest {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int T = 1;
        int[] ans = new int[10000+1];
        while (T-- > 0) {
            int t = sc.nextInt();
            int[] n = new int[t];
            int[] r = new int[t];
            Arrays.fill(ans, -1);
            for (int i = 0; i < t; i++) n[i] = sc.nextInt();

            for (int i = 0; i < t; i++) {
                r[i] = sc.nextInt();

            }
            int mod = (int)(1e9+7);
            for( int i = 0; i < t ;i++ ) {
                if( ans[i] != -1 ) out.println(ans[i]);
                else {
                    long anss = pow(2, r[i], mod);
                    ans[i] = (int)anss%mod;
                    out.println(ans[i]);
                }
            }

        }
    }
    public static long pow( int x, int a, int mod ) {
        if( a == 0 ) return 1;
        int y =(int) pow( x, a/2, mod) % mod ;
        if((a&1) == 0 ) return ((long)y*y) % mod;
        return (((long)x*((long)y*y) % mod) % mod);
    }
}

**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 buggy code, we need to focus on the following aspects:

1. **Edge Cases**: Values of `n` and `k` that are at the boundaries of the allowed range.
2. **Non-Edge Cases**: Values of `n` and `k` that are in the middle of the allowed range.
3. **Large Values**: Values of `n` and `k` that are large but within the allowed range to test the performance and correctness of the algorithm.
4. **Specific Patterns**: Values of `n` and `k` that might cause the faulty algorithm to produce incorrect results.

Given the buggy code, the primary issue is that it uses a simple power of 2 calculation to determine the binomial coefficients, which is incorrect according to the problem's requirements. The correct binomial coefficient calculation should follow the formula `C[n][k] = C[n - 1][k] + C[n - 1][k - 1]`.

### Test Input

Let's create a small-scale test input that includes a mix of edge cases and non-edge cases to ensure the buggy code fails.

```plaintext
5
2 5 100000 100000 100000
1 2 1 33333 66666
```

### Explanation

1. **(2, 1)**: This is a small and simple case. The correct binomial coefficient is 2, but the buggy code will output 2, which is correct by coincidence.
2. **(5, 2)**: This is a slightly more complex case. The correct binomial coefficient is 10, but the buggy code will output 4, which is incorrect.
3. **(100000, 1)**: This is an edge case for `k`. The correct binomial coefficient is 100000, but the buggy code will output 2, which is incorrect.
4. **(100000, 33333)**: This is a large value for `k`. The correct binomial coefficient is much larger, but the buggy code will output a large power of 2, which is incorrect.
5. **(100000, 66666)**: This is another large value for `k`. The correct binomial coefficient is much larger, but the buggy code will output a large power of 2, which is incorrect.

This test input should effectively expose the bug in the buggy code by showing that the output is incorrect for most of the cases.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
int main() {
    long long t;
    cin >> t;
    vector<long long int> n(t);
    vector<long long int> k(t);
    for(int i=0; i<t;i++) cin>>n[i];
    for(int i=0; i<t;i++) cin>>k[i];
    for(int i=0;i<t;i++){
        long long ans = 1;
        long long power = k[i];
        while (power--) {
            ans = (ans * 2) % MOD;
        }
        cout << ans << endl;
    }

    return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided buggy code, we need to focus on edge cases and scenarios where the incorrect formula produces significantly different results compared to the correct formula. The key issue in the buggy code is that it calculates \( C[n][k] \) as \( 2^k \) instead of using the correct binomial coefficient formula.

### Steps to Generate the Test Input:
1. **Edge Cases:**
   - Small values of \( n \) and \( k \) to ensure the results are manageable and easy to verify.
   - Values of \( k \) close to \( n \) to maximize the difference between the correct and incorrect formulas.
   - Large values of \( n \) and \( k \) to test the modulo operation.

2. **Input Format:**
   - The first line contains the number of test cases \( t \).
   - The second line contains the values of \( n \) for each test case.
   - The third line contains the values of \( k \) for each test case.

### Test Input:
```plaintext
5
2 3 4 100 100000
1 2 3 50 99999
```

### Explanation:
- **Test Case 1:**
  - \( n = 2 \), \( k = 1 \)
  - Correct: \( C[2][1] = 2 \)
  - Buggy: \( 2^1 = 2 \)
- **Test Case 2:**
  - \( n = 3 \), \( k = 2 \)
  - Correct: \( C[3][2] = 3 \)
  - Buggy: \( 2^2 = 4 \)
- **Test Case 3:**
  - \( n = 4 \), \( k = 3 \)
  - Correct: \( C[4][3] = 4 \)
  - Buggy: \( 2^3 = 8 \)
- **Test Case 4:**
  - \( n = 100 \), \( k = 50 \)
  - Correct: \( C[100][50] \) (a large number)
  - Buggy: \( 2^{50} \) (a large number, but different from the correct binomial coefficient)
- **Test Case 5:**
  - \( n = 100000 \), \( k = 99999 \)
  - Correct: \( C[100000][99999] \) (a large number)
  - Buggy: \( 2^{99999} \) (a large number, but different from the correct binomial coefficient)

This test input should effectively expose the bug in the buggy code by showing significant differences between the expected and actual results.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

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

int main()
{
    int t;
    cin>>t;

    int n[t], k[t];
    for(int i=0; i<t; i++)  cin>>n[i];
    for(int i=0; i<t; i++)  cin>>k[i];

    for(int i=0; i<t; i++)
    {
        long long ans = 1;
        int mod = 1000000007;
        if(k[i]==0 || k[i]==n[i])
        ans = 1;

        else
        for(int j= 0; j<k[i]; j++)
        {
            ans = (ans%mod)*2;
        }

        ans = ans%mod;

        cout<<ans<<endl;

    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, let's consider the following points:

1. **Edge Cases**:
   - The smallest possible values for \( n \) and \( k \).
   - The largest possible values for \( n \) and \( k \).
   - Values where \( k \) is close to \( n \).

2. **Behavior of the Buggy Code**:
   - The buggy code calculates \( C[n][k] \) as \( 2^k \) for \( k > 0 \) and \( k < n \).
   - This is incorrect for the actual binomial coefficient formula, but we need to focus on the behavior of the given code.

3. **Input Constraints**:
   - \( 1 \le t \le 10^5 \)
   - \( 2 \le n_i \le 10^5 \)
   - \( 1 \le k_i < n_i \)

Given these points, a small-scale test input that includes a variety of cases can be:

```plaintext
4
2 3 100000 100000
1 2 1 99999
```

### Explanation:
- **Case 1**: \( n = 2, k = 1 \)
  - According to the buggy code, \( C[2][1] = 2 \).
- **Case 2**: \( n = 3, k = 2 \)
  - According to the buggy code, \( C[3][2] = 4 \).
- **Case 3**: \( n = 100000, k = 1 \)
  - According to the buggy code, \( C[100000][1] = 2 \).
- **Case 4**: \( n = 100000, k = 99999 \)
  - According to the buggy code, \( C[100000][99999] = 2^{99999} \mod 1000000007 \).

This test input covers small and large values of \( n \) and \( k \), and includes a case where \( k \) is very close to \( n \). It should effectively expose the incorrect behavior of the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
MOD = 10**9 + 7
t = int(input())
n = list(map(int, input().split()))
k = list(map(int, input().split()))
for i in range(t):
    if k[i] == 0 or k[i] == n[i]:
        print(1)
    else:
        result = pow(2, k[i])
        result = result % MOD
        print(result)

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

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

Think step by step.

Response:
To expose the bug in the given buggy code, we need to focus on scenarios where the incorrect formula `C[n][k] = C[n][k - 1] + C[n - 1][k - 1]` produces significantly different results compared to the correct formula `C[n][k] = C[n - 1][k] + C[n - 1][k - 1]`.

### Steps to Generate the Test Input:
1. **Edge Cases**:
   - Small values of `n` and `k` where the difference is noticeable.
   - Large values of `n` and `k` to test the performance and correctness under modulo operation.
   - Cases where `k` is close to `n` to see if the incorrect formula diverges significantly.

2. **Small-Scale Test Input**:
   - We will choose a small number of test cases, say 5, to ensure the input is manageable.
   - Include a mix of small and large values for `n` and `k`.

### Test Input:
```plaintext
5
3 4 5 10 100
1 2 3 5 99
```

### Explanation:
- **Case 1**: `n = 3, k = 1`
  - Correct formula: `C[3][1] = C[2][1] + C[2][0] = 2 + 1 = 3`
  - Buggy formula: `C[3][1] = C[3][0] + C[2][0] = 1 + 1 = 2`
- **Case 2**: `n = 4, k = 2`
  - Correct formula: `C[4][2] = C[3][2] + C[3][1] = 3 + 3 = 6`
  - Buggy formula: `C[4][2] = C[4][1] + C[3][1] = 2 + 3 = 5`
- **Case 3**: `n = 5, k = 3`
  - Correct formula: `C[5][3] = C[4][3] + C[4][2] = 4 + 6 = 10`
  - Buggy formula: `C[5][3] = C[5][2] + C[4][2] = 5 + 6 = 11`
- **Case 4**: `n = 10, k = 5`
  - Correct formula: `C[10][5] = C[9][5] + C[9][4] = 126 + 126 = 252`
  - Buggy formula: `C[10][5] = C[10][4] + C[9][4] = 21 + 126 = 147`
- **Case 5**: `n = 100, k = 99`
  - Correct formula: `C[100][99] = C[99][99] + C[99][98] = 1 + 9900 = 9901`
  - Buggy formula: `C[100][99] = C[100][98] + C[99][98] = 2^99 + 9900 = 633825300114114700748351602688 % 1000000007 = 633825300114114700748351602688 % 1000000007 = 326186014`

This test input will help expose the bug in the buggy code by showing significant differences in the results produced by the incorrect formula.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
t = int(input())
n = [int(j) for j in input().split()]
k = [int(z) for z in input().split()]
for i in range(t):
    ans = (2 ** k[i]) % 1000000007
    print(ans)

**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 buggy code, we need to focus on the specific behavior of the faulty formula. The faulty formula given is:

```plaintext
C[n][k] = C[n][k - 1] + C[n - 1][k - 1]
```

This formula is incorrect because it does not correctly compute the binomial coefficients. Instead, it computes a different sequence. The correct formula is:

```plaintext
C[n][k] = C[n - 1][k] + C[n - 1][k - 1]
```

The faulty formula tends to produce values that are powers of 2 for certain inputs. Specifically, for \( k = 1 \), the faulty formula will produce \( 2 \) for any \( n \geq 1 \). For larger \( k \), the values can grow exponentially.

To expose the bug, we should choose inputs where the difference between the correct and faulty formulas is most apparent. Here are the steps to generate a small-scale test input:

1. **Choose a small number of pairs \( t \) to keep the input manageable.**
2. **Select \( n_i \) and \( k_i \) values that highlight the exponential growth in the faulty formula.**
3. **Include edge cases where \( k_i \) is close to \( n_i \) to see the difference in large values.**

Let's generate the test input:

```plaintext
5
2 5 10 100 1000
1 2 3 10 500
```

### Explanation:
- **First pair (2, 1):**
  - Correct: \( C[2][1] = 2 \)
  - Faulty: \( C[2][1] = 2 \)
- **Second pair (5, 2):**
  - Correct: \( C[5][2] = 10 \)
  - Faulty: \( C[5][2] = 4 \)
- **Third pair (10, 3):**
  - Correct: \( C[10][3] = 120 \)
  - Faulty: \( C[10][3] = 8 \)
- **Fourth pair (100, 10):**
  - Correct: \( C[100][10] = 17310309456440 \)
  - Faulty: \( C[100][10] = 1024 \)
- **Fifth pair (1000, 500):**
  - Correct: \( C[1000][500] \) is a very large number
  - Faulty: \( C[1000][500] = 2^{500} \mod 1000000007 \)

This test input will clearly show the difference between the correct and faulty formulas, especially for larger values of \( k \) and \( n \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
import java.util.*;
import java.io.*;
import java.math.*;
public class Main
{
  public static void main(String[]args) throws IOException
  {
    BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
    PrintWriter pw=new PrintWriter(System.out);
    StringTokenizer st=new StringTokenizer(br.readLine());
    int t=Integer.parseInt(st.nextToken());
    String[]a=br.readLine().split(" ");
    String[]b=br.readLine().split(" ");
    for(int i=0;i<t;i++)
    {
      long x=Integer.parseInt(a[i]);
      long y=Integer.parseInt(b[i]);
      long mod=1000000007;
      if(x==y)
      pw.println(1);
      else
      {
        BigInteger n=BigInteger.valueOf(2);
        n=n.pow((int)y);
        pw.println(n.mod(BigInteger.TEN.pow(9).add(BigInteger.valueOf(7))));
      }
    }
    pw.close();
  }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
import java.util.Scanner;

public class BinomialCoefficientsKindOf {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[n];
        int[] b = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
        }
        for (int i = 0; i < n; i++) {
            b[i] = sc.nextInt();
        }
        for (int i = 0; i < n; i++) {
            long ans = 2;
            for (int j = 1; j < b[i]; j++) {
                ans = (ans * 2) % 1000000007;
            }
            System.out.println(ans);
        }
    }
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided buggy code, we need to focus on the specific issue in the code. The buggy code calculates the binomial coefficients using a simplified formula that essentially computes powers of 2. The correct formula for binomial coefficients is more complex and involves summing previous values.

The bug in the code is that it uses `ans = (ans * 2) % 1000000007` to compute the coefficients, which is incorrect for the problem's requirements. To expose this, we need to provide inputs where the correct binomial coefficient is significantly different from the power of 2.

Let's consider the following test input:

1. A small number of pairs to keep the input size manageable.
2. Pairs where the correct binomial coefficient is not a power of 2.
3. Pairs where the correct binomial coefficient is significantly different from the power of 2.

Here is a suitable test input:

```plaintext
3
5 10 15
2 5 7
```

### Explanation:
1. **Pair (5, 2)**:
   - Correct binomial coefficient: \( C[5][2] = 10 \)
   - Buggy code output: \( 2^2 = 4 \)

2. **Pair (10, 5)**:
   - Correct binomial coefficient: \( C[10][5] = 252 \)
   - Buggy code output: \( 2^5 = 32 \)

3. **Pair (15, 7)**:
   - Correct binomial coefficient: \( C[15][7] = 6435 \)
   - Buggy code output: \( 2^7 = 128 \)

This test input will clearly show the discrepancy between the correct binomial coefficients and the values produced by the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
const long long mod = 1e9+7;

int t,a;
long long s;

int main(){
	cin>>t;
	for(int i=1;i<=t;i++) cin>>a;
	for(int i=1;i<=t;i++){
		cin>>s;
		long long res = 1;
		for(int j=1;j<=s;j++){
			res = (res*2)%mod;
		}
		cout<<res<<endl;
	}
	return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on the following aspects:

1. **Understanding the Bug**: The buggy code calculates the binomial coefficients using a simple power of 2, which is not the correct method according to the problem's requirements. The correct formula should be:
   \[
   C[n][k] = C[n - 1][k] + C[n - 1][k - 1]
   \]
   However, the buggy code uses:
   \[
   C[n][k] = C[n][k - 1] + C[n - 1][k - 1]
   \]

2. **Edge Cases and Large Values**: The test input should include edge cases and large values to maximize the failure probability. This will help us verify if the code can handle such cases correctly.

3. **Small-Scale Test Input**: The test input should be small enough to fit within the constraints but still be challenging enough to expose the bug.

Let's construct a small-scale test input that includes a mix of small and large values for \( n \) and \( k \):

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

### Explanation:
- **Test Case 1**: \( n = 2 \), \( k = 1 \)
  - According to the correct formula, \( C[2][1] = C[1][1] + C[1][0] = 1 + 1 = 2 \).
  - The buggy code will calculate \( 2^1 = 2 \), which happens to be correct in this case.

- **Test Case 2**: \( n = 5 \), \( k = 2 \)
  - According to the correct formula, \( C[5][2] = C[4][2] + C[4][1] \).
    - \( C[4][2] = 6 \)
    - \( C[4][1] = 4 \)
    - \( C[5][2] = 6 + 4 = 10 \)
  - The buggy code will calculate \( 2^2 = 4 \), which is incorrect.

- **Test Case 3**: \( n = 10 \), \( k = 5 \)
  - According to the correct formula, \( C[10][5] = C[9][5] + C[9][4] \).
    - \( C[9][5] = 126 \)
    - \( C[9][4] = 126 \)
    - \( C[10][5] = 126 + 126 = 252 \)
  - The buggy code will calculate \( 2^5 = 32 \), which is incorrect.

This test input should effectively expose the bug in the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

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

const int modulo = 1e9 + 7; // Correcting the modulo constant

int main() {
    int t;
    cin >> t;

    vector<int> n(t);
    vector<int> k(t);
    vector<int> ans(t); // This will hold the final results

    // Read n values
    for (int i = 0; i < t; i++) {
        cin >> n[i]; // Use n[i] instead of n[t]
    }

    // Read k values
    for (int i = 0; i < t; i++) {
        cin >> k[i]; // Use k[i] instead of k[t]
    }

    // Calculate answers
    for (int i = 0; i < t; i++) {
        ans[i] = 1; // Initialize ans[i] to 1

        for (int j = 1; j <= k[i]; j++) {
            ans[i] = (ans[i] * 2) % modulo; // Update ans[i] correctly
        }

        cout << ans[i] << endl; // Print the answer
    }

    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 3 4
1 2 3
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
temp = int(input())
x = list(map(int, input().split()))
y = list(map(int, input().split()))

for i in range(len(x)):
    print(pow(2,y[i]) % 1000000007)

**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 scenarios where the incorrect formula produces significantly different results compared to the correct formula. The buggy code calculates \( C[n][k] \) as \( 2^k \) modulo \( 10^9 + 7 \), which is clearly different from the correct binomial coefficient calculation.

### Steps to Generate the Test Input:
1. **Choose a small number of test cases**: Since the output length is limited, we should choose a small number of test cases, say 5.
2. **Select values of \( n \) and \( k \)**: We should choose values of \( n \) and \( k \) such that the difference between the correct binomial coefficient and \( 2^k \) is maximized. This can happen for larger values of \( k \) and \( n \).

### Chosen Test Cases:
- \( (n_1, k_1) = (5, 3) \): The correct binomial coefficient \( C[5][3] = 10 \), but the buggy code will output \( 2^3 = 8 \).
- \( (n_2, k_2) = (10, 5) \): The correct binomial coefficient \( C[10][5] = 252 \), but the buggy code will output \( 2^5 = 32 \).
- \( (n_3, k_3) = (20, 10) \): The correct binomial coefficient \( C[20][10] = 184756 \), but the buggy code will output \( 2^{10} = 1024 \).
- \( (n_4, k_4) = (50, 25) \): The correct binomial coefficient \( C[50][25] \) is a large number, but the buggy code will output \( 2^{25} = 33554432 \).
- \( (n_5, k_5) = (100, 50) \): The correct binomial coefficient \( C[100][50] \) is a very large number, but the buggy code will output \( 2^{50} = 1125899906842624 \).

### Test Input:
```plaintext
5
5 10 20 50 100
3 5 10 25 50
```

This test input will clearly show the discrepancy between the correct binomial coefficients and the values produced by the buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
t=int(input())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
for i in range(t):
    if a[i]==b[i]:
        print(1)
    else:
        print((2**(b[i]))%1000000007)

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

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

Think step by step.

Response:
To expose the bug in the provided buggy code, we need to focus on the differences between the correct binomial coefficient formula and the incorrect one used in the code. The incorrect formula is:

\[ C[n][k] = C[n][k - 1] + C[n - 1][k - 1] \]

The correct formula is:

\[ C[n][k] = C[n - 1][k] + C[n - 1][k - 1] \]

The key difference is that the incorrect formula uses \( C[n][k - 1] \) instead of \( C[n - 1][k] \). This will lead to different values, especially for larger \( n \) and \( k \).

To create a challenging test input, we should choose values of \( n \) and \( k \) that highlight this difference. Specifically, we should choose values where \( k \) is significantly less than \( n \), and \( n \) is large enough to show the discrepancy clearly.

Let's consider the following test input:

```plaintext
5
10 20 50 100 1000
5 10 25 50 500
```

Explanation:
- \( t = 5 \) (number of pairs)
- The pairs are:
  - \( (n_1, k_1) = (10, 5) \)
  - \( (n_2, k_2) = (20, 10) \)
  - \( (n_3, k_3) = (50, 25) \)
  - \( (n_4, k_4) = (100, 50) \)
  - \( (n_5, k_5) = (1000, 500) \)

These pairs are chosen to ensure that \( k \) is significantly less than \( n \), which will highlight the difference between the correct and incorrect formulas. The values are also large enough to show the impact of the incorrect formula on the final result.

This test input will help expose the bug in the provided buggy code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

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

    public static void solve(Scanner sc){
        int mod=1000000007;
        int t=sc.nextInt();
        ArrayList<Integer> arr1=new ArrayList<>(t);
        ArrayList<Integer> arr2=new ArrayList<>(t);
        for(int i=0;i<t;i++){
            arr1.add(sc.nextInt());
        }
        for(int i=0;i<t;i++){
            arr2.add(sc.nextInt());
        }
        ArrayList<Integer> ans=new ArrayList<>(t);
        int i=0;
        while(i<t){

            int num1= arr1.get(i);
            int num2=arr2.get(i);
            if(num2==0 || num2==num1){
                ans.add(1);
            }
            else{
                int ct=1;
                while(num2-->0){
                    ct=(ct*2)%mod;
                }
                ans.add(ct);
            }
            i++;
        }
        for(int j=0;j<t;j++){
            System.out.println(ans.get(j));
        }
    }
    public static void main(String[] arg){
        Scanner sc=new Scanner(System.in);
        int test=1;
//        int test=sc.nextInt();
        while(test-->0){
            solve(sc);
        }
    }
}

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

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

Think step by step.

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


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Binomial Coefficients, Kind Of
time_limit_ms: 2000
memory_limit_mb: 512
Description: Recently, akshiM met a task that needed binomial coefficients to solve. He wrote a code he usually does that looked like this:

                  for (int n = 0; n < N; n++) { // loop over n from 0 to N-1 (inclusive)               C[n][0] = 1;               C[n][n] = 1;               for (int k = 1; k < n; k++) // loop over k from 1 to n-1 (inclusive)                   C[n][k] = C[n][k - 1] + C[n - 1][k - 1];           }

Unfortunately, he made an error, since the right formula is the following:

                          C[n][k] = C[n - 1][k] + C[n - 1][k - 1]

But his team member keblidA is interested in values that were produced using the wrong formula. Please help him to calculate these coefficients for $t$ various pairs $(n_i, k_i)$. Note that they should be calculated according to the first (wrong) formula.

Since values $C[n_i][k_i]$ may be too large, print them modulo $10^9 + 7$.

Input: The first line contains a single integer $t$ ($1 \le t \le 10^5$) — the number of pairs. Next, $t$ pairs are written in two lines.

The second line contains $t$ integers $n_1, n_2, \dots, n_t$ ($2 \le n_i \le 10^5$).

The third line contains $t$ integers $k_1, k_2, \dots, k_t$ ($1 \le k_i < n_i$).

Output: Print $t$ integers $C[n_i][k_i]$ modulo $10^9 + 7$.

Examples:
input:
7
2 5 5 100000 100000 100000 100000
1 2 3 1 33333 66666 99999
output:
2
4
8
2
326186014
984426998
303861760

Note:

Buggy Code:
import java.util.*;

public class Main {
    public static void main(String[] args) {
      Scanner sc=new Scanner(System.in);
      //System.out.println(Integer.MAX_VALUE);
      int t=sc.nextInt();
      int[] arr=new int[t];
      for(int i=0;i<t;i++)
        arr[i]=sc.nextInt();
      for(int i=0;i<t;i++)
        arr[i]=sc.nextInt();
      for(int i=0;i<t;i++)
        System.out.println(fun(arr[i]));
    }
    public static int fun(int k){
      int ans=1;
      for(int i=1;i<=k;i++){
        ans*=2;
        ans%=1000000007;
      }
      return ans;
    }
}

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

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

Think step by step.

Response:
To create a challenging test input that exposes the bug in the provided buggy code, we need to focus on the following aspects:

1. **Edge Cases**: Values of `n` and `k` that are at the boundaries of the allowed ranges.
2. **Large Values**: Values that are large enough to potentially cause overflow or other issues.
3. **Special Cases**: Values that might cause the incorrect formula to produce different results compared to the correct formula.

Given the constraints:
- \(1 \le t \le 10^5\)
- \(2 \le n_i \le 10^5\)
- \(1 \le k_i < n_i\)

We will create a small-scale test input with a few pairs that cover these aspects.

### Step-by-Step Reasoning:
1. **Small `t`**: Use a small number of pairs to keep the input manageable.
2. **Small `n` and `k`**: Use small values to ensure the input is valid and easy to verify.
3. **Large `n` and `k`**: Use large values to test the behavior with large inputs.
4. **Edge Cases**: Use values at the boundaries of the allowed ranges.

### Test Input:
```plaintext
3
2 100000 100000
1 1 99999
```

### Explanation:
- **Pair 1**: \(n = 2\), \(k = 1\)
  - This is a small pair that should be easy to compute and verify.
  - According to the buggy formula, \(C[2][1] = C[2][0] + C[1][0] = 1 + 1 = 2\).

- **Pair 2**: \(n = 100000\), \(k = 1\)
  - This is a large pair with \(k = 1\).
  - According to the buggy formula, \(C[100000][1] = C[100000][0] + C[99999][0] = 1 + 1 = 2\).

- **Pair 3**: \(n = 100000\), \(k = 99999\)
  - This is a large pair with \(k\) just below \(n\).
  - According to the buggy formula, \(C[100000][99999] = C[100000][99998] + C[99999][99998]\).
  - The result will be a large number, and we need to ensure it is computed correctly modulo \(10^9 + 7\).

This test input will help expose the bug in the provided buggy code, especially in the way it handles large values and edge cases.