**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

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

using namespace std;

#define ll long long
#define vll vector<long long>
#define vpl vector<pair<long long, long long>>
#define _fastio  ios_base:: sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define lmin LONG_LONG_MIN
#define lmax LONG_LONG_MAX
#define mxn 5004

ll t, n, m, k, mod = 1e9 + 7, ans;
string s, s1;

ll ncr[5001][5001];
ll nCr(ll i, ll r) {
    if(ncr[i][r]!=-1) return ncr[i][r];
    if(r==0) return ncr[i][r]=1ll;
    if(i<r) return ncr[i][r]=0ll;
    return ncr[i][r]=(nCr(i-1,r-1)+nCr(i-1,r))%mod;
}

signed main()
{
    _fastio;
    cin >> t;
    while (t--)
    {
        cin >> n;
        ll ans = 0;
        memset(ncr, -1, sizeof(ncr));
        for (ll i = 1; i <= n; i++)
        {
            for (ll z = i + 1; z <= 2 * i + 1; z++)
            {
                if (n - i + max(0ll, z - n) < i + 1)
                {
                    continue;
                }

                ll left = nCr(min(n, z - 1), z - i - 1), right = 1;
                if (n >= z)
                {
                    right = nCr(n - z, 2 * i - z + 1);
                }
                // cout << i << ' ' << z << ' ' << left << ' ' << right << ' ' << '\n';
                ans += (((left * right) % mod) * z) % mod;
                ans %= mod;
            }
        }
        cout << ans + 1 << '\n';
    }

    return 0;
}

/*
1 3
2 3
3 2
4 2
5 2

1 2
*/

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

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

Think step by step.

Response:
```plaintext
5
1
2
3
4
5000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

Buggy Code:
// LUOGU_RID: 166129792
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define pii pair<int,int>
#define endl '\n'
const int N=2e5+5;
int n,m;
const int P = 1e9+7;
int *fac, *ifac;
int qpow(int a, int b, int p = P)
{
    int res = 1;
    for (; b; b >>= 1, a = a * a % p)
    {
        if (b & 1)
        {
            res = res * a % p;
        }
    }
    return res;
}
int inv(int x)
{
    return qpow(x, P - 2, P);
}
void init(int N)
{
    fac = new int [N + 1];//½×³Ë
    ifac = new int [N + 1];//½×³ËÄæÔª
    for (int i = 0; i <= N; i++)
    {
        fac[i] = ifac[i] = 0;
    }
    fac[0] = 1;
    for (int i = 1; i <= N; i++)
    {
        fac[i] = fac[i - 1] * i % P;
    }
    ifac[N] = inv(fac[N]);
    for (int i = N - 1; i >= 0; i--)
    {
        ifac[i] = ifac[i + 1] * (i + 1) % P;
    }
}

int comb(int n, int m)
{
    if (m < 0 || m > n || n < 0)
    {
        return 0;
    }
    return fac[n] * ifac[m] % P * ifac[n - m] % P;
}
void solve()
{
    cin>>n;
    int ans=1;
    init(1e5);
    for(int i=1; i<=n; i++)
    {
        if(n>2*i)
        {
            ans+=comb(n-(i+1),i)*(i+1);
            for(int j=1; j<i; j++)
            {
                ans+=comb(j+i,j)%P*comb(n-j-i-1,i-j)%P*(j+i+1)%P;
                //cout<<j<<" "<<comb(2*i-1,j)*comb(n-2*i,i-j)%P*(j+i+1)<<endl;
                ans%=P;

            }
            ans+=comb(2*i,i)*(2*i+1);
            //cout<<ans<<endl;
        }
        else ans+=(2*i+1)*comb(n,i)%P;
        //cout<<ans<<endl;
    }
    cout<<ans%P<<endl;
}
signed main()
{
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);
    std::cout.tie(0);
    int _=1;
    cin>>_;
    while(_--)
    {
        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To generate a test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that might cause the code to fail. The code involves combinatorial calculations and modular arithmetic, so we should consider the following:

1. **Small values of \( n \)**: These are straightforward and less likely to expose bugs.
2. **Large values of \( n \)**: These can stress the combinatorial and modular arithmetic functions.
3. **Special values of \( n \)**: Values that might cause boundary conditions to be tested, such as \( n = 1 \) or \( n = 2 \).

Given the constraints and the nature of the problem, a good test case to expose potential bugs would be a value of \( n \) that is large enough to stress the combinatorial calculations but still within the problem's limits.

Let's choose \( n = 5000 \) as it is the maximum value for \( n \) and will stress the combinatorial and modular arithmetic functions.

Here is the test input:

```plaintext
1
5000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

Buggy Code:
from math import comb
from time import time, time_ns

def main():
    debug = False
    show = False
    show_more = False

    M = 10 ** 9 + 7

    tests = 1 if debug else int(input())
    ns = [int(input()) for _ in range(tests)]
    ts = time()
    n_max = max(ns)
    if show:
        print(tests, ns, n_max)

    # mex_amount = [[0 for j in range(0, n_max + 1)] for i in range(n_max + 1)]
    # for i in range(0, n_max + 1):
    #     mex_amount[0][i] = 1
    #     for j in range(i, n_max + 1):
    #         mex_amount[i][j] = comb(j, i)

    # mex_value = [[2 * i + 1 if j >= i else 0 for j in range(0, n_max + 1)] for i in range(n_max + 1)]

    # mex_value = [[2 * i + 1 if j >= i else 0 for j in range(0, n_max + 1)] for i in range(n_max + 1)]

    if show_more:
        print("mex_amount")
        for i in range(n_max + 1):
            for j in range(n_max + 1):
                print(f"{mex_amount[i][j]:4}", end=' ')
            print()

    if show_more:
        print("mex_value")
        for i in range(n_max + 1):
            for j in range(n_max + 1):
                print(f"{mex_value[i][j]:4}", end=' ')
            print()

    corrections = [0 for _ in range(n_max + 1)]
    corrections[3] = -1
    corrections[5] = -8

    for test in range(tests):

        n = ns[test]
        # a = [i for i in range(1, n+1)]

        if n == 4999:
            print(354226409)
            continue

        s = 0
        if show:
            print(f"{test = } {n = }")
        for i in range(0, n + 1):
            # последний элемент = comb(n_max, i)
            # сумма в этйо строке равна comb(n_max+1, i+1)
            # значение в этой строке равно max_value[i][n_max]
            num = comb(n, i)
            value = 2 * i + 1
            ss = ((num % M) * value) % M
            s = (s + ss) % M
            if show:
                print(f"\ncomb(n, i) = {num = } {value = }")
                print(f"{i = } {n = } {s = } {ss = } {corrections[n]}")

        s += corrections[n]
        print(s % M)
        if show_more:
            print(f"{time() - ts:5.4f}")

main()

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

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

Think step by step.

Response:
To expose the bug in the provided code, we need to focus on edge cases and specific values that might cause the algorithm to fail. The code has a known issue with specific values of `n` (e.g., `n = 3` and `n = 5`), where it applies manual corrections. However, there might be other values of `n` that are not explicitly handled and could still cause issues.

Let's consider the following steps to generate a challenging test input:

1. **Edge Cases**: Test with the smallest and largest possible values of `n` (i.e., 1 and 5000).
2. **Known Corrections**: Test with values of `n` that have known corrections (i.e., 3 and 5).
3. **Intermediate Values**: Test with intermediate values to see if the algorithm behaves correctly without manual corrections.

Given the constraints and the need for a small-scale test input, we can create a test case that includes these values.

Here is the test input:

```plaintext
5
1
3
5
10
5000
```

This test input includes:
- `n = 1` (smallest possible value)
- `n = 3` (known correction)
- `n = 5` (known correction)
- `n = 10` (intermediate value)
- `n = 5000` (largest possible value)

This set of test cases should help expose any bugs in the algorithm, especially in the handling of specific 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: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

Buggy Code:
import java.util.Scanner;
import java.util.HashSet;
import java.util.Set;

public class UltraMeow {
    private static final int MOD = 1000000007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        while (t-- > 0) {
            int n = sc.nextInt();
            System.out.println(calculateMEOW(n));
        }
        sc.close();
    }

    private static long calculateMEOW(int n) {
        if (n == 1) return 4;
        if (n == 2) return 12;
        if (n == 3) return 31;
        if (n == 5) return 184;
        if (n == 4999) return 354226409;

        long sum = 0;
        for (int mask = 0; mask < (1 << n); mask++) {
            Set<Integer> subset = new HashSet<>();
            for (int i = 0; i < n; i++) {
                if ((mask & (1 << i)) != 0) {
                    subset.add(i + 1);
                }
            }
            sum = (sum + MEX(subset, subset.size() + 1)) % MOD;
        }
        return sum;
    }

    private static int MEX(Set<Integer> set, int k) {
        int mex = 1;
        while (k > 0) {
            if (!set.contains(mex)) {
                k--;
            }
            mex++;
        }
        return mex - 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:
```plaintext
5
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: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

Buggy Code:
#include <bits/stdc++.h>
#define mp make_pair
#define ll long long
#define migmig cin.tie(NULL);ios_base::sync_with_stdio(false)
#define max_heap(T) priority_queue<T>
#define min_heap(T) priority_queue<T, vector<T>, greater<T>>
#define pb push_back
#define fi first
#define sec second
#define endl "\n"
#define pii pair <int, int>
using namespace std;
const ll MOD = 1e9 + 7;
const int INF = 1e9;
const ll INF18 = 1e18;
const int MAXN = 1e5 + 1;
const int LOG = 23;
ll fact[MAXN];
ll inv[MAXN];
ll mul_mod(ll a, ll b) {return (a * b) % MOD;}
ll pow_mod(ll a, ll b){
	ll ans;
	if (b == 0) return 1;
	if (b % 2 == 0) ans = pow_mod(a, b / 2), ans = mul_mod(ans, ans);
	else ans = mul_mod(a, pow_mod(a, b - 1));
	return ans % MOD;
}
ll c(ll n, ll r){
    if (r > n || r < 0) return 0;
    return mul_mod(mul_mod(fact[n], inv[r]), inv[n - r]);
}
void solve(){
    fact[0] = 1;
    inv[0] = 1;
    for (int i = 1; i < MAXN; i++){
        fact[i] = fact[i - 1] * i;
        fact[i] %= MOD;
        inv[i] = pow_mod(fact[i], MOD - 2);
    }
    ll n;
    cin >> n;
    ll t = 0;
    for (int i = 0; i < (n + 1) / 2; i++){
        for (int j = 1; j <= n; j++){
            t += (c(j - 1, i) * c(n - j, 2 * i - j + 1) % MOD) * j % MOD;
            t %= MOD;
        }
    }
    for (int i = (n + 1) / 2; i <= n; i++) t+= c(n, i) * (2 * i + 1) % MOD, t %= MOD;
    cout << t << endl;
}
int main(){
	migmig;
	int tc = 1;
	cin >> tc;
	for (int ti = 0; ti < tc; ti++) solve();
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that can cause the code to fail. Let's analyze the code and identify potential issues:

1. **Factorial and Inverse Calculation:**
   - The factorial and inverse calculations are done up to `MAXN = 1e5 + 1`, which is much larger than the maximum `n` of 5000. This part seems fine.

2. **Combinatorial Calculations:**
   - The code uses combinatorial calculations (`c(n, r)`) to compute the MEX values. The combinatorial calculations involve large numbers, and the modulo operations are necessary to prevent overflow.

3. **Summation and Modulo Operations:**
   - The summation and modulo operations are performed in a nested loop. The loop structure and the way the summation is handled can be a source of bugs, especially for large `n`.

4. **Edge Cases:**
   - Small values of `n` (e.g., `n = 1` and `n = 2`) are simple and less likely to expose bugs.
   - Large values of `n` (e.g., `n = 5000`) can be challenging due to the large number of combinatorial calculations and potential overflow issues.

Given these considerations, let's generate a test input that can expose the bug:

- **Small `n` with a known result:**
  - `n = 3` is a good candidate because it is small but not trivial. The expected result for `n = 3` is `31`.

- **Medium `n` with a known result:**
  - `n = 5` is another good candidate. The expected result for `n = 5` is `184`.

- **Large `n` to test the limits:**
  - `n = 4999` is a large value that can stress the combinatorial calculations and modulo operations. The expected result for `n = 4999` is `354226409`.

Here is the test input:

```plaintext
3
3
5
4999
```

This test input includes a small, medium, and large value of `n` to thoroughly test the code and expose any potential bugs.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

Buggy Code:
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
const int N=5e3+5;
const int EN=1e9+7;
ll c[N][N];
void clean(){
	for(int i=0;i<N;i++){
		c[i][0]=1;
	}
	for(int i=1;i<N;i++){
		for(int j=1;j<=i;j++){
			c[i][j]=(c[i-1][j]+c[i-1][j-1])%EN;
		}
	}
}
int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	int t;
	cin>>t;
	while(t--){
		int n;
		clean();
		ll ans=0;
		cin>>n;
		for(int i=0;i<=n/2-(n%2==0);i++){
			for(int j=i+1;j<=i+i+1;j++){
				ans+=(1ll*j*c[j-1][j-i-1]%EN*c[n-j][i-(j-i-1)]%EN);
				ans%=EN;
			}
		}
		for(int i=n/2+(n%2==1);i<=n;i++){
			ans+=(1ll*(i+i+1)*c[n][i]%EN);
			ans%=EN;
		}
		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:
To generate a challenging test input that exposes the bug in the given buggy code, we need to focus on edge cases and scenarios that are likely to cause issues. Here are the steps to think through this:

1. **Edge Cases:**
   - **Small Arrays:** Test with the smallest possible array sizes to ensure basic functionality.
   - **Large Arrays:** Test with the largest possible array sizes to ensure the algorithm handles large inputs efficiently.
   - **Odd and Even Sizes:** Test with both odd and even-sized arrays to ensure the code handles both cases correctly.
   - **Special Values:** Test with values that might cause unexpected behavior, such as when `n` is a power of 2 or when `n` is a prime number.

2. **Specific Scenarios:**
   - **Single Element Array:** Test with `n = 1` to ensure the code handles the simplest case.
   - **Two Element Array:** Test with `n = 2` to ensure the code handles small arrays correctly.
   - **Large Prime Number:** Test with a large prime number to ensure the code handles large, non-power-of-2 values.
   - **Large Power of 2:** Test with a large power of 2 to ensure the code handles large, power-of-2 values.

3. **Combining Scenarios:**
   - Combine multiple test cases to ensure the code can handle a variety of inputs in a single run.

Based on these considerations, a challenging test input could be:

```plaintext
5
1
2
3
4999
1024
```

This test input includes:
- The smallest possible array size (`n = 1`).
- A small array size (`n = 2`).
- A small odd-sized array (`n = 3`).
- A large array size close to the upper limit (`n = 4999`).
- A large power of 2 (`n = 1024`).

These test cases are designed to cover a range of edge cases and scenarios that could expose bugs 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: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

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

#define endl '\n'
#define INF 0x3f3f3f3f
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
const ll mod = 1e9+7;
const int N = 5000 + 30;
const int M = 1010;
#define int long long

int C[N][N],dp[N];

int CC(int a,int b){
    if(a<0||b<0||a<b){
        return 0;
    }
    else {
        return C[a][b];
    }
}

void work() {
    int n;
    cin>>n;
    C[0][0]=1;
    for(int i=1;i<=5000;i++){
        C[i][0]=1;
        C[i][i]=1;
        for(int j=1;j<=i;j++){
            C[i][j]=(C[i-1][j]+C[i-1][j-1])%mod;
        }
    }

    for(int i=1;i<=5000;i++){
        for(int j=0;j<=i;j++){
            if((2*j+1)<=i){
                continue;
            }
            else{
                dp[i]+=(C[i][j]*(2*j+1))%mod;
            }
            dp[i]%=mod;
        }
    }
    for(int i=1;i<=n;i++){
        int x;
        cin>>x;
        for(int j=0;2*j+1<=x;j++){
            int num=0;
            for(int k=j+1;k<=2*j+1;k++){
                dp[x]+=k%mod*CC(k-1,num)%mod*CC(x-k+1-1,j-num)%mod;
                dp[x]%=mod;
                num++;
            }
        }
        cout<<dp[x]<<endl;
    }
}

signed main () {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cout.tie(nullptr);
    int t;
    t = 1;
    //cin>>t;
    while (t --)
        work();
    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 for the given buggy code, we need to focus on edge cases and scenarios that are likely to expose the bug. Here are the steps to think through this:

1. **Edge Cases**:
   - **Smallest Possible Input**: `n = 1`
   - **Largest Possible Input**: `n = 5000`
   - **Mid-range Input**: `n = 2500`

2. **Special Cases**:
   - **Small Inputs with Specific Patterns**: `n = 2`, `n = 3`, `n = 4`
   - **Inputs that might cause overflow or edge behavior in the combinatorial calculations**: `n = 5000`

3. **Behavior Analysis**:
   - The code calculates the MEX values and uses combinatorial functions, which can be complex and prone to errors for large inputs.
   - The code also uses a dynamic programming approach, which can be sensitive to initial conditions and boundary cases.

Given these considerations, a small-scale test input that is complete and valid and likely to expose the bug is:

```plaintext
3
1
2
5000
```

This test input includes:
- The smallest possible input (`n = 1`), which is a simple case.
- A small input (`n = 2`), which is a common edge case for combinatorial problems.
- The largest possible input (`n = 5000`), which is a more complex and resource-intensive case.

This combination should help identify any issues in the combinatorial calculations and dynamic programming logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

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

using namespace std;

#define ll long long int
#define vi vector<ll>
#define vvi vector<vi>
#define vb vector<bool>
#define rep(i, n) for (ll i = 0; i < n; i++)
#define nl cout << endl
#define all(x) x.begin() , x.end()
#define read(a,n) rep(i,n)cin>>a[i]
#define give(a,n) rep(i,n)cout<<a[i]<<" "
#define pii pair<ll,ll>
#define pb push_back
#define f first
#define s second
const ll inf = 1e18;
const ll mod = 1e9+7;
const ll mod1 = 998244353  ;
const ll maxn = 5005 ;
vi dp(maxn , 1) ;

ll inv(ll x){
    x = x%mod ;
    ll y = mod-2 , ans = 1  ;
    while(y){
        if(y&1) ans = (ans*x)%mod ;
        x = (x*x)%mod ;
        y = y>>1 ;
    }
    return ans ;
}

ll nck(ll n , ll k ){
    if(n<0 || k<0 || n<k) return 0 ;
    return ((dp[n]*inv(dp[k]*dp[n-k]))%mod) ;
}

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

    for(ll i = 1 ; i<maxn ; i++) dp[i] = (dp[i-1]*i)%mod ;
    ll t = 1; cin >> t;
    while (t--){
        ll n ; cin>>n;
        ll ans = 1 ;
        for(ll len = 1; len<=n ;  len++){
            if(len+len+1 > n ) ans= (ans + nck(n,len)*(2*len+1))%mod  ;
            else{
                for(ll mex = len+1 ; mex <= 2*len + 1 ; mex++){
                    ll temp = (nck(mex-1 , mex-len-1)*nck(n-mex , 2*len+1-mex))%mod ;
                    ans = (ans + mex*temp)%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:
```plaintext
5
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: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

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

using namespace std;
#define int long long

const int mod = 1e9 + 7, maxn = 2e5 + 5;

int f[maxn];

void factorize(){
    f[0] = 1;
    for(int i = 1; i < maxn; i++) f[i] = (f[i - 1] * i) % mod;
}

int binpow(int a, int b, int m){
    a %= m;
    int res = 1;
    while(b > 0){
        if(b & 1) res = (res * a) % m;
        a = (a * a) % m;
        b /= 2;
    }
    return res;
}

int Cnk(int n, int k){
    return f[n] * binpow(f[k] * f[n - k], mod - 2, mod) % mod;
}

void solve(){
    //duyệt qua tất cả độ dài của subset và mex của nó
    int n; cin >> n;

    int ans = 1;//subset độ dài 0
    for (int k = 1; k <= n; k++) {
        //2 * k + 1 sẽ là phần tử mex
        //phần tử mex chắc chắn sẽ lớn hơn n
        if (2 * k >= n) {
            ans = (ans + Cnk(n, k) * (2 * k + 1)) % mod;
            continue;
        }

        //lúc này mex sẽ nằm trong k + 1 tới 2 * k + 1
        //có chính xác m - k - 1 số trước m và 2 * k + 1 số sau m
        //vậy số cách chọn subset độ dài k để có mex là m là
        for (int m = k + 1; m <= 2 * k + 1; m++) {
            int c = Cnk(m - 1, k) * Cnk(n - m, 2 * k + 1 - m) % mod;
            ans = (ans + (c * m % mod)) % mod;
        }
    }

    cout << ans << '\n';

}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);

    factorize();
    int t = 1;
    cin >> t;
    while(t--) solve();

    //cout << (7 ^ 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
1
5000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

Buggy Code:
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#define endl "\n"
#define fo(i, n) for (i = 0; i < n; i++)
#define Fo(i, k, n) for (i = k; k < n; k++)
#define pb push_back
#define ll long long
#define mp make_pair
#define ff first
#define ss second
#define all(x) x.begin(), x.end()
#define clr(x) memset(x, 0, sizeof(x))
#define sortall(x) sort(all(x))
#define tr(it, arr) for (auto it = arr.begin(); it != arr.end(); it++)
#define PI 3.1415926535897932384626
#define suprit                    \
    ios_base::sync_with_stdio(0); \
    cout.tie(0);                  \
    cin.tie(0);
#define line cout << endl;

typedef pair<int, int> pi;
typedef pair<ll, ll> pl;
typedef vector<int> vi;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<vi> vvi;
typedef vector<vl> vvl;
typedef map<ll, ll> ml;
typedef map<string, ll> msl;
typedef map<ll, string> mls;
typedef unordered_map<ll, ll> uml;
typedef unordered_map<string, ll> umsl;
typedef unordered_map<ll, string> umls;
typedef set<ll> sl;
typedef set<pair<ll, ll>> spl;
typedef ordered_set<ll> osl;
typedef ordered_set<pair<ll, ll>> ospl;

const ll mod = 1e9 + 7;

bool comp2(pair<ll, ll> &arr, pair<ll, ll> &b)
{
    if (arr.first == b.first)
        return arr.second < b.second;
    return arr.first < b.first;
};

template <typename T>
void read(T i, T n, vector<T> &arr)
{
    for (T j = i; j < n; j++)
        cin >> arr[j];
}
template <typename T>
void read(vector<T> &arr)
{
    for (auto &j : arr)
        cin >> j;
}

#ifndef ONLINE_JUDGE

template <typename T, typename V>
void print(set<pair<T, V>> &arr)
{
    for (auto &it : arr)
        cout << it.first << " " << it.second << endl;
    line
}
template <typename T, typename V>
void print(pair<T, V> &pa) { cout << pa.first << " " << pa.second << endl; }
template <typename T>
void print(T i, T last, vector<T> &arr)
{
    for (T j = i; j < last; j++)
        cout << arr[j] << " ";
    line
}
template <typename T>
void print(T i, vector<T> &arr)
{
    for (T j = i; j < arr.size(); j++)
        cout << arr[j] << " ";
    line
}
template <typename T>
void print(vector<T> &arr)
{
    for (auto &i : arr)
        cout << i << " ";
    line
}
template <typename T, typename V>
void print(unordered_map<T, V> &arr)
{
    for (auto &it : arr)
        cout << it.first << " " << it.second << endl;
    line
}
template <typename T, typename V>
void print(map<T, V> &arr)
{
    for (auto &it : arr)
        cout << it.first << " " << it.second << endl;
}
template <typename T>
void print(unordered_set<T> &arr)
{
    for (auto &it : arr)
        cout << it << " ";
    line
}
template <typename T>
void print(ordered_set<T> &arr)
{
    for (auto &it : arr)
        cout << it << " ";
    line
}
template <typename T>
void print(set<T> &arr)
{
    for (auto &it : arr)
        cout << it << " ";
    line
}
template <typename T, typename... Args>
void print(T t, Args... args)
{
    cout << t << " ";
    print(args...);
}
template <typename T>
void print(T t) { cout << t << "\n"; }
#define debug(x) cout << #x << " " << x << endl;
#define debug2(x, y) cout << #x << " " << x << " " << #y << " " << y << endl;
#define debug3(x, y, z) cout << #x << " " << x << " " << #y << " " << y << " " << #z << " " << z << endl;
#define debug4(x, y, z, a) cout << #x << " " << x << " " << #y << " " << y << " " << #z << " " << z << " " << #a << " " << a << endl;

#else

#define print(x)
#define debug(x)
#define debug2(x, y)
#define debug3(x, y, z)
#define debug4(x, y, z, a)

#endif

// 5 5 4 3 2 1
// (n-5) c 5
// 4 c 1
const int nmax = 10002;
vl factorial(nmax), inv_fact(nmax);
map<ll, ll> already_calc;

long long binpow(long long a, long long b, long long m)
{
    a %= m;
    long long res = 1;
    while (b > 0)
    {
        if (b & 1)
            res = res * a % m;
        a = a * a % m;
        b >>= 1;
    }
    return res;
}

void precalc()
{
    factorial[0] = 1;
    for (ll i = 1; i <= nmax; i++)
        factorial[i] = factorial[i - 1] * i % mod;
    inv_fact[nmax - 1] = binpow(factorial[nmax - 1], mod - 2, mod);
    for (ll i = nmax - 2; i >= 0; i--)
        inv_fact[i] = inv_fact[i + 1] * (i + 1) % mod;
}

ll nCr(ll n, ll r)
{
    if (r == 0 || r == n)
        return 1;
    if (r > n || r < 0)
        return 0;
    ll ans = factorial[n] * inv_fact[r] % mod;
    ans = ans * inv_fact[n - r] % mod;
    return ans;
}

void solve()
{
    ll n;
    cin >> n;
    // if (already_calc.count(n))
    // {
    //     cout << already_calc[n] << endl;
    //     return;
    // }
    ll ans = 1;
    for (ll i = 2; i <= n; i++)
    {
        for (ll k = 1; k <= i; k++)
        {
            if(i - k > k - 1) continue;
            ll temp = nCr(i - 1, i - k) * nCr(n - i, 2 * k - i - 1) % mod;
            temp = temp * i % mod;
            // debug3(i, k, temp);
            ans = (ans + temp) % mod;
        }
    }
    for(ll i = n+1; i <= 2*n+1; i++) {
        if(i % 2 == 0) continue;
        ll temp = i * nCr(n, i/2) % mod;
        ans = (ans + temp) % mod;
    }
    // already_calc[n] = ans;
    cout << ans << endl;
}

int main()
{
    suprit;
    clock_t start = clock();

    int t = 1;
    precalc();
    cin >> t;
    while (t--)
    {
        solve();
    }
    clock_t end = clock();

#ifndef ONLINE_JUDGE
    double elapsed = double(end - start) / CLOCKS_PER_SEC;
    cout << setprecision(10) << elapsed << endl;
#endif
    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
5
1
2
3
4
5000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

Buggy Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define endl "\n"

const int M=1e9+7;

int power(int a, int b, int m) {
    int ans = 1, x = a;
    while (b) {
        if (b & 1) ans = (ans * x) % m;
        x = (x * x) % m;
        b /= 2;
    }
    return ans;
}

int divide(int a, int b, int m=M) {
    int b0 = power(b, m - 2, m);
    return (a * b0) % m;
}

int fac[200010];
int nCm(int n,int r){
    return (divide(fac[n],(fac[r]*fac[n-r])%M))%M;
}

signed main() {
    ios_base::sync_with_stdio(false); cin.tie(NULL);
    fac[0]=1;
    for (int i=1;i<200010;i++) fac[i]=(i*fac[i-1])%M;
    int t;
    cin>>t;
    while(t--){
        int n,ans=0;
        cin>>n;
        for(int a=0;a<=n;a++){
            if(2*a>=n){
                ans=(ans+nCm(n,a)*(2*a+1))%M;
            }
            else{
                ans+=nCm(n,a)*(a+1)%M;
                for(int i=1; i<=a; i++){
                    int xxx=a+i;
                    if(n-xxx-1<0) break;
                    ans=(ans+nCm(xxx,i)*nCm(n-xxx-1,a-i)%M*i%M)%M;
                }
            }
        }
        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:
```plaintext
1
4999
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

Buggy Code:
#include <iostream>
using namespace std;
const int MAXN = 2e5 + 5;
const int mod = 1e9 + 7;
int n;
long long f[MAXN], ans;
long long qpow(long long a, int b) {
    long long ret = 1;
    while (b) {
        if (b & 1) {
            ret = ret * a % mod;
        }
        a = a * a % mod;
        b >>= 1;
    }
    return ret;
}
long long c(int n, int m) {
    long long p = f[n];
    long long q = f[m] * f[n - m] % mod;
    return p * qpow(q, mod - 2) % mod;
}

int main() {
    int T;
    scanf("%d", &T);
    f[0] = 1;
    for (int i = 1; i < MAXN; i++) {
        f[i] = f[i - 1] * i % mod;
    }
    while (T--) {
        scanf("%d", &n);
        ans = 1;
        for (int k = 1; k <= n; k++) {
            if (k * 2 >= n) {
                ans = (ans + c(n, k) * (2 * k + 1) % mod) % mod;
            } else {
                for (int m = k + 1; m <= k * 2 + 1; m++) {
                    ans = (ans + (c(m - 1, m - 1 - k) *
                                  c(n - m, 2 * k + 1 - m) % mod) *
                                     m % mod) %
                          mod;
                }
            }
        }
        printf("%lld\n", ans);
    }
    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 that can highlight potential issues. The code involves combinatorial calculations and modular arithmetic, so we should consider inputs that can cause issues in these areas.

### Step-by-Step Reasoning:

1. **Edge Case: Small Array Size**:
   - Test with the smallest possible array size, `n = 1`.
   - This is a simple case but can help verify the base behavior.

2. **Edge Case: Large Array Size**:
   - Test with the largest possible array size, `n = 5000`.
   - This can help identify issues with large combinatorial calculations and potential overflow or performance issues.

3. **Edge Case: Middle Range Array Size**:
   - Test with a medium-sized array, `n = 2500`.
   - This can help identify issues that might not be apparent with very small or very large inputs.

4. **Edge Case: Specific Values**:
   - Test with specific values that might trigger edge conditions in the combinatorial calculations, such as `n = 3` and `n = 4`.

5. **Multiple Test Cases**:
   - Include multiple test cases to ensure the code handles various inputs correctly.

### Chosen Test Input:

```plaintext
5
1
2
3
2500
5000
```

### Explanation:

- **Test Case 1 (n = 1)**:
  - This is the smallest possible array, which should be straightforward to handle.
  - Expected output: 4 (since the only subset is the array itself, and the MEX is 2).

- **Test Case 2 (n = 2)**:
  - A small array to check basic combinatorial calculations.
  - Expected output: 12 (as given in the example).

- **Test Case 3 (n = 3)**:
  - A slightly larger array to check more complex combinatorial calculations.
  - Expected output: 31 (as given in the example).

- **Test Case 4 (n = 2500)**:
  - A medium-sized array to check performance and correctness.
  - Expected output: A large number (to be verified).

- **Test Case 5 (n = 5000)**:
  - The largest possible array to check the limits of the combinatorial calculations and potential overflow.
  - Expected output: A large number (to be verified).

This test input should help expose any bugs in the provided code by covering a range of array sizes and combinatorial scenarios.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

Buggy Code:
mod = 10**9 + 7
n = 100000
fact = [1 for i in range(n+1)]
fact_inv = [1 for i in range(n+1)]
for i in range(1,n+1):
  fact[i] = fact[i-1]*i % mod
fact_inv[-1] = pow(fact[-1],mod-2,mod)
for i in range(n,0,-1):
  fact_inv[i-1] = fact_inv[i]*i % mod
def binom(n,r):
  if n < r or n < 0 or r < 0:
    return 0
  res = fact_inv[n-r] * fact_inv[r] % mod
  res *= fact[n]
  res %= mod
  return res

for _ in range(int(input())):
    n = int(input())
    ans = 1
    for le in range(1,n+1):
        for v in range(le+1,n+1):
            res = binom(v-1,le) * binom(n-v,2*le-v+1) % mod
            res = res * v % mod
            ans += res
            ans %= mod
        # over
        if 2*le + 1 > n:
            v = 2*le + 1
            ans += v * binom(n,le) % mod
            ans %= mod
    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 provided code, we need to focus on edge cases and scenarios that maximize the failure probability. Here are the steps to consider:

1. **Edge Cases:**
   - **Small Values of \( n \):** Test with \( n = 1 \) and \( n = 2 \) to ensure the base cases are handled correctly.
   - **Large Values of \( n \):** Test with \( n = 5000 \) to ensure the algorithm handles large inputs efficiently and correctly.

2. **Boundary Conditions:**
   - **Middle Values of \( n \):** Test with \( n = 3 \) and \( n = 5 \) to ensure the algorithm works for typical intermediate values.

3. **Specific Scenarios:**
   - **Values that Trigger Specific Conditions:** Test with values that might trigger specific conditions in the code, such as \( n = 4999 \).

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

```plaintext
5
1
2
3
5
5000
```

This test input includes:
- The smallest possible value \( n = 1 \).
- A small value \( n = 2 \).
- A medium value \( n = 3 \).
- Another medium value \( n = 5 \).
- The largest possible value \( n = 5000 \).

This should help expose any bugs in the code, especially in handling different ranges of \( n \).


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

Buggy Code:
#include "bits/stdc++.h"
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;

// #include <boost/multiprecision/cpp_int.hpp>
// using namespace boost::multiprecision;

// #include <atcoder/all>
// using namespace atcoder;

using namespace std;

// ==== Load debug template ====
#ifndef ONLINE_JUDGE
#include </home/srinivas/Srinivas/CP/templates/debug.hpp>
#define debug(x...)               \
    cerr << "[" << #x << "] = ["; \
    _print(x)
#else
#define debug(x...)
#endif
// ==== End of debug template ====

// ================================== take ip/op like vector,pairs directly!==================================
template <typename typC, typename typD>
istream &operator>>(istream &cin, pair<typC, typD> &a) {return cin >> a.first >> a.second;}
template <typename typC>
istream &operator>>(istream &cin, vector<typC> &a) { for (auto &x : a) cin >> x; return cin;}
template <typename typC, typename typD>
ostream &operator<<(ostream &cout, const pair<typC, typD> &a) { return cout << a.first << ' ' << a.second;}
template <typename typC, typename typD>
ostream &operator<<(ostream &cout, const vector<pair<typC, typD>> &a) { for (auto &x : a) cout << x << '\n'; return cout; }
template <typename typC>
ostream &operator<<(ostream &cout, const vector<typC> &a) {
    int n = a.size();
    if (!n) return cout;
    cout << a[0];
    for (int i = 1; i < n; i++) cout << ' ' << a[i];
    return cout;
}
// ===================================END Of the input module ==========================================

#define all(c) (c).begin(),(c).end()
#define endl "\n"
#define ff first
#define ss second
#define allr(c) (c).rbegin(),(c).rend()
#define ifr(i, begin, end) for (__typeof(end) i = (begin) - ((begin) > (end)); i != (end) - ((begin) > (end)); i += 1 - 2 * ((begin) > (end)))
#define pof pop_front
#define pob pop_back
#define pb emplace_back
#define pf emplace_front
#define fstm(m,n,r) m.reserve(n);m.max_load_factor(r)
#define mp make_pair
#define mt make_tuple
#define inf LLONG_MAX
#define os tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
//order_of_key (k) : Number of items strictly smaller than k .
//find_by_order(k) : K-th element in a set (counting from zero).
const double PI = acos(-1);
typedef complex<double> cd;
typedef long long ll;
ll gcd() { return 0ll; } template<typename T, typename... Args> T gcd(T a, Args... args) { return __gcd(a, (__typeof(a))gcd(args...)); }
typedef map<int, int> mii;
typedef map<ll, ll> mll;
typedef unordered_map<ll, ll> umll;
typedef unordered_map<int, int> umii;
typedef vector<ll> vll;
typedef vector<int> vi;
typedef vector<vector<ll>> vvll;
typedef vector<vector<int>> vvi;
typedef pair<ll, ll> pll;
typedef pair<int, int> pii;
typedef long double ld;
#define mod 1000000007

const int N = 5e3 + 100;

// source :- https://codeforces.com/blog/entry/63903

template <int MOD=1000000007>
struct Modular {
  int value;
  static const int MOD_value = MOD;

  Modular(long long v = 0) { value = v % MOD; if (value < 0) value += MOD;}
  Modular(long long a, long long b) : value(0){ *this += a; *this /= b;}

  Modular& operator+=(Modular const& b) {value += b.value; if (value >= MOD) value -= MOD; return *this;}
  Modular& operator-=(Modular const& b) {value -= b.value; if (value < 0) value += MOD;return *this;}
  Modular& operator*=(Modular const& b) {value = (long long)value * b.value % MOD;return *this;}

  friend Modular power(Modular a, long long e) {
    Modular res = 1; while (e) { if (e&1) res *= a; a *= a; e >>= 1; }
    return res;
  }
  friend Modular inverse(Modular a) { return power(a, MOD - 2); }

  Modular& operator/=(Modular const& b) { return *this *= inverse(b); }
  friend Modular operator+(Modular a, Modular const b) { return a += b; }
  friend Modular operator-(Modular a, Modular const b) { return a -= b; }
  friend Modular operator-(Modular const a) { return 0 - a; }
  friend Modular operator*(Modular a, Modular const b) { return a *= b; }
  friend Modular operator/(Modular a, Modular const b) { return a /= b; }
  friend std::ostream& operator<<(std::ostream& ostr, Modular const& a) {return ostr << a.value;}
  friend std::istream& operator>>(std::istream& istr, Modular &a) {return istr >> a.value;}
  friend bool operator==(Modular const& a, Modular const& b) {return a.value == b.value;}
  friend bool operator!=(Modular const& a, Modular const& b) {return a.value != b.value;}
};

typedef Modular<mod> mint;

mint fact[N];

mint ncr(int n, int r) {
  if(r>n) return 0;
  return fact[n] / (fact[n-r] * fact[r]);
}

void Solve() {
  int n;
  cin>>n;

  mint res = 1, s = 0;
  int l, r;
  for(int j=2; j<=n; j++) {
    l = (j-1)>>1;
    r = !(j&1);
    s = 0;

    while(r<=n-j && l>=0) {
      s += ncr(j-1, l) * ncr(n-j, r);
      r+=2;
      l--;
    }
    res += j * s;
  }

  for(int j=n+1; j<=2*n+1; j++) {
    l = (j-1)>>1;
    if(j&1) {
      res += j * ncr(n, l);
    }
  }

  cout<<res<<endl;
}

int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  // #ifndef ONLINE_JUDGE
  //     freopen("input.txt","r",stdin);
  // #endif

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

  int T;cin >> T;while (T--)
  {
    Solve();
  }
  return 0;
}

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

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

Think step by step.

Response:
```plaintext
5
1
2
3
4
5000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

Buggy Code:
#include <iostream>
#include <vector>
#include <set>
#include <algorithm>
#include <string>
#include <queue>
#include <numeric>
#include <map>
#include <stack>
#include <math.h>
#include <unordered_map>
#include <deque>
#include <stdlib.h>
#include <string.h>
#include <random>
#include <chrono>

#ifndef ONLINE_JUDGE
#include "debug.h"
#else
#define debug(x)
#endif

using namespace std;

typedef long long ll;

ll MOD = 1e9 + 7;
const int N = (int) 5e3 + 10;
int INF = 1e9 + 5;

vector<int> factorial (2 * N, 0);
vector<int> inverse_factorial(2 * N, 0);
map<int, ll> res;

ll add(const ll& x, const ll& y) {
    return (x + y) % MOD;
}

ll substract(const ll& x, const ll& y) {
    return(x - y + MOD) % MOD;
}

ll mul(const ll& x, const ll& y) {
    return (x * y) % MOD;
}

ll inverse(const ll& x) {
    return x <= 1 ? x : MOD - (ll)(MOD / x) * inverse(MOD % x) % MOD;
}

ll div(const ll& x, const ll& y) {

}

void preprocess() {
    factorial[0] = 1;
    inverse_factorial[0] = 1;
    for (int i = 1; i < N; i++) {
        factorial[i] = mul(i, factorial[i - 1]);
        inverse_factorial[i] = inverse(factorial[i]);
    }
}

int combinaison(int k, int n) { // choose k in n
    if (n < 0 || k < 0 || n < k) return 0;
    if (n == 0 || k == 0) return 1;
    return mul(mul(factorial[n], inverse_factorial[k]), inverse_factorial[n - k]);
}

void solve(int test_case) {
    int n; cin >> n;

    if (res.find(n) != res.end()) {
        printf("%lld\n", res[n]);
        return;
    }

    ll ans = 1; // empty subset

    for (int k = 1; k < n; k++) {
        if (2 * k >= n) {
            ans += mul(combinaison(k, n), (2 * k + 1));
            continue;
        }

        // all values from k + 1 to 2K + 1
        for (int m = k + 1; m <= 2 * k + 1; m++) {
            ans = add(ans, mul(mul(m, combinaison(m - k - 1, m - 1)), combinaison(2 * k + 1 - m, n - m)));
        }
    }

    ans = add(ans, 2 * n + 1); // full subset
    res[n] = ans;

    printf("%lld\n", ans);
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t = 1;
    cin >> t;
    for (int i = 1; i <= t; i++) {
        preprocess();
        solve(i);
    }
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given code, we need to focus on edge cases and scenarios that can highlight issues in the algorithm. The code provided has several potential pitfalls, such as incorrect handling of combinatorial calculations and modular arithmetic.

Here are the steps to generate a challenging test input:

1. **Edge Cases**: Test with the smallest and largest possible values of \( n \) to ensure the code handles boundary conditions correctly.
2. **Special Cases**: Test with \( n \) values that are powers of 2, as these can often reveal issues in combinatorial calculations.
3. **Intermediate Values**: Test with intermediate values to ensure the code handles general cases correctly.

Given the constraints \( 1 \le t \le 10^4 \) and \( 1 \le n \le 5000 \), a small-scale test input that covers these cases effectively would be:

```plaintext
5
1
2
5000
2500
3
```

This input includes:
- \( n = 1 \) (smallest possible value)
- \( n = 2 \) (a small value to test basic combinatorial logic)
- \( n = 5000 \) (largest possible value)
- \( n = 2500 \) (a large intermediate value)
- \( n = 3 \) (a small intermediate value)

This test input is designed to cover a range of possible values and edge cases, which should help in exposing any bugs in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

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

public class G1992 {

    static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    static long[] fact , r;
    static long mod9= 1000000007l;
    public static void main(String[] args)throws IOException{
                int inputs = Integer.parseInt(br.readLine());
                for(int i=0 ; i<inputs ; i++){
                           solve();
                }
    }
     public static Integer readInteger()throws IOException{

              Integer res = Integer.parseInt(br.readLine());
              return res;
    }

     public static int[] readInt(int n)throws IOException{

              String[] h =  br.readLine().split(" ");
              int[] res = new int[n];
              for(int i=0 ; i<n ; i++){
                        res[i] = Integer.parseInt(h[i]);
              }
              return res;
    }

     public static long readLong()throws IOException{

              long res = Long.parseLong(br.readLine());
              return res;
    }

    public static long[] readLong(int n)throws IOException{

              String[] h =  br.readLine().split(" ");
              long[] res = new long[n];
              for(int i=0 ; i<n ; i++){
                        res[i] = Long.parseLong(h[i]);
              }
              return res;
    }
    public static void solve()throws IOException{
        int n = readInteger();
        long ans=0;
        fact = new long[15000];
        r = new long[15000];
        fact[0] = 1l;
        for(int i=1 ; i<fact.length ; i++){
            fact[i] = mul(fact[i-1], i);
        }

        int N = fact.length;
        r[N - 1] = inv(fact[N - 1]);
        for (int i = N - 2; i > -1; i--)
            r[i] = mul(r[i + 1], i + 1);

        // r[0] = 1;
        // for(int i=1 ; i<N ; i++){
        //     r[i] = mul(r[i-1] , inv(4));
        // }

        for(long k=0 ; k<=n ; k++){
            if(n<=2*k){
                ans = add(ans,mul(choose(n, k),(2*k+1)));
                continue;
            }

            for(long m=k+1 ; m<= 2*k+1 ; m++){
                ans = add(ans, mul( mul(choose(m-1, m-1-k),choose(n-m,2*k+1-m)) ,m) );
            }
        }
        System.out.println(ans);
}

    public static long choose(long n , long k){
        if(n<k || n<0 || k<0)return 0;
        return mul(fact[(int) n], (mul( r[(int) (n-k)], r[(int) k] )) );
    }

    public static long mul(long a , long b){
        return (a*b)%mod9;
    }

    public static long add(long a , long b){
        if (a + b >= mod9)
            return a + b - mod9;
        return a + b;
    }

    public static long sub(long a , long b){
        if (a < b)
            return a + mod9 - b;
        return a - b;
    }
    public static long inv(long a) {
        return a <= 1 ? a : mod9 - (long)(mod9/a) * inv(mod9 % a) % mod9;
    }

    static long power(long a , long b){
            if(b==0) return a;
            long res = power(a , b/2);
            if(b%2==1)
              return res * res * a;
            else
              return res * res;
        }

}

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

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

Think step by step.

Response:
```plaintext
1
5000
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

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

// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>

#define int long long
#define double long double
#define input(a) for(auto &x:a) cin>>x
#define all(x) x.begin(),x.end()
#define rsort(c) sort(all(c)); reverse(all(c))
#define print(a) for(auto x:a) cout<<x<<" "; cout<<endl
#define print1(a) for(auto x:a) cout<<x.first<<" "<<x.second<<endl
#define printall(a) for(auto x:a){print(x);} cout<<endl
#define fil(ar,val) memset(ar,val,szof(ar))
#define mpp make_pair
#define pb push_back
#define fi first
#define se second
// const int MOD=998244353;
const int MOD=1e9+7;

typedef vector<int>vi;
typedef vector<vector<int>>vvi;
typedef pair<int,int>pi;
typedef vector<pair<int,int>> vpi;
typedef vector<bool>vb;
typedef pair<int, int> pi;

template<typename T> using PQ = priority_queue<T>;
template<typename T> using QP = priority_queue<T,vector<T>,greater<T>>;
template<typename T>T min(const vector<T>&v){return *min_element(v.begin(),v.end());}
template<typename T>T max(const vector<T>&v){return *max_element(v.begin(),v.end());}
template<typename T>T acc(const vector<T>&v){return accumulate(v.begin(),v.end(),T(0));};

// typedef __gnu_pbds::tree<int, __gnu_pbds::null_type, less<int>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update> ordered_set;
// order_of_key: The number of items in a set that are strictly smaller than k

#ifndef ONLINE_JUDGE
#define debug(x) cerr<<#x<<" ";_print(x); cerr<<endl;
#else
#define debug(x)
#endif

void _print(int a){cerr<<a;}
void _print(char a){cerr<<a;}
void _print(string a){cerr<<a;}
void _print(bool a){cerr<<a;}
void _print(pair<int,int>a){cerr<<'['<<a.first<<" "<<a.second<<']';}

template<class T> void _print(vector<T> v1){
    cerr<<"[ ";
    for(T i:v1){
        _print(i);
        cerr<<" ";
    }
    cerr<<"]";
}

template<class T> void _print(set<T> s){
    cerr<<"[ ";
    for(T i:s){
        _print(i);
        cerr<<" ";
    }
    cerr<<"]";
}

struct mi {
    int v; explicit operator int() const { return v; }
    mi() { v = 0; }
    mi(long long _v) : v(_v % MOD) { v += (v < 0) * MOD; }
};
mi& operator+=(mi& a, mi b) {
    if ((a.v += b.v) >= MOD) a.v -= MOD;
    return a;
}
mi& operator-=(mi& a, mi b) {
    if ((a.v -= b.v) < 0) a.v += MOD;
    return a;
}
mi operator+(mi a, mi b) { return a += b; }
mi operator-(mi a, mi b) { return a -= b; }
mi operator*(mi a, mi b) { return mi((long long) a.v * b.v); }
mi& operator*=(mi& a, mi b) { return a = a * b; }
mi pow(mi a, long long p) {
    assert(p >= 0);
    return p == 0 ? 1 : pow(a * a, p / 2) * (p & 1 ? a : 1);
}
mi inv(mi a) { assert(a.v != 0); return pow(a, MOD - 2); }
mi operator/(mi a, mi b) { return a * inv(b); }
bool operator==(mi a, mi b) { return a.v == b.v; }
bool operator==(mi a, int b){ return a.v == b; }
ostream& operator<<(ostream& os, const mi& val){ os << (int) val; return os; }

//this part is for binomial coefficients
const int maxn=2e5+5;
mi fact[maxn],ifact[maxn];
void ncr(){
    fact[0]=fact[1]=1;
    for(int i=2;i<maxn;i++){
        fact[i]=fact[i-1]*i;
    }
    ifact[maxn-1]=inv(fact[maxn-1]);
    for(int i=maxn-2;i>=0;i--){
        ifact[i]=(i+1)*ifact[i+1];
    }
}
mi ncr(int n,int r){
    if(n==0) return 1;
    if(n<r) return 0;
    return fact[n]*ifact[n-r]*ifact[r];
}

void solve(){
    int n;
    cin>>n;

    ncr();
    mi ans=0;
    for(int i=1;i<=2*n+1;i++){
        //in how many subsets is the MEX = i
        mi curr=0;
        int r=0;
        if(i>n) r=i-n-1;
        for(int j=max(r,(i-1)/2);j<=(i-1);j++){
            if(2*j+1>=i){
                mi x = ncr(min(i-1,n),j-r);
                // cerr<<j<<"\n";
                int need=(2*j+1-i);//this is the extra elements needed
                if(i>=n&&need){
                    x=0;
                }
                else if(i<n){
                    if(need<=(n-i)) x*=ncr(n-i,need);
                    else x=0;
                }
                // cerr<<x<<"\n\n";
                curr+=x;
            }
        }
        curr*=i;
        ans+=curr;
    }

    cout<<ans<<"\n";
}

//this method is not perfect i guess

signed main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin>>t;

    while(t--){

        solve();
    }
    return 0;
}

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

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

Think step by step.

Response:
To expose the bug in the given code, let's focus on a few key aspects:

1. **Edge Cases**: Small values of \( n \) can help identify issues with the logic, especially when \( n = 1 \) or \( n = 2 \).
2. **Boundary Conditions**: Values of \( n \) that are near the upper limit of the problem constraints can help identify issues with large inputs.
3. **Specific Patterns**: Inputs that might cause unexpected behavior in the code, such as values where \( n \) is a power of 2 or other specific patterns.

Given the constraints and the nature of the problem, a small-scale test input that can expose potential issues is:

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

This input includes a variety of small values for \( n \) which can help identify issues with the logic in the `solve` function. Let's break down why this input is chosen:

1. **\( n = 1 \)**: This is the smallest possible input, and it should be straightforward to compute. Any issues here can indicate fundamental problems with the logic.
2. **\( n = 2 \)**: This is the next smallest input, and it can help identify issues with handling small sets.
3. **\( n = 3 \)**: This is a slightly larger input, which can help identify issues with handling more complex sets.
4. **\( n = 4 \)**: This is a moderate-sized input, which can help identify issues with the combinatorial calculations.
5. **\( n = 5 \)**: This is a larger input, which can help identify issues with the combinatorial calculations and the handling of larger sets.

By using this input, we can test the code's ability to handle a range of small values and identify any potential bugs in the logic.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

Buggy Code:
class xzlMath( object ):
    f = None
    rf = None
    N = 5000
    MOD = 1000000007

    @staticmethod
    def lcm(a : int, b : int):
        return a * b // xzlMath.gcd(a, b)

    @staticmethod
    def gcd(a : int, b : int):
        if b == 0:
            return a
        else:
            return xzlMath.gcd(b, a % b)

    @staticmethod
    def qp(x: int, q: int = None):
        if q is None: q = xzlMath.MOD - 2
        MOD = xzlMath.MOD
        res = 1
        while q > 0:
            if (q & 1) > 0:
                res = res * x % MOD
            q >>= 1
            x = x * x % MOD
        return res

    @staticmethod
    def C(n: int, m: int):
        if n < 0 or m > n or m < 0:
            return 0
        if xzlMath.f is None:
            xzlMath.f = [1 for _ in range(xzlMath.N + 1)]
            xzlMath.rf = [1 for _ in range(xzlMath.N + 1)]
            for i in range(1, xzlMath.N + 1):
                xzlMath.f[i] = xzlMath.f[i - 1] * i % xzlMath.MOD
            xzlMath.rf[xzlMath.N - 1] = xzlMath.qp(xzlMath.f[xzlMath.N - 1])
            for i in range(xzlMath.N - 1, 0, -1):
                xzlMath.rf[i - 1] = xzlMath.rf[i] * i % xzlMath.MOD
        f = xzlMath.f
        rf = xzlMath.rf
        MOD = xzlMath.MOD
        return f[n] * rf[m] % MOD * rf[n - m] % MOD

def sol():
    n = int(input())
    ans = 0
    for i in range(1, 2 * n + 2):
        for b in range(n + 1):
            pre_in = i - 1 - b
            if pre_in < 0: break
            ls_black = max(0, n - i)
            ls_in = b - pre_in
            ans += xzlMath.C(min(n, i - 1), pre_in) * xzlMath.C(ls_black, ls_in) * i % xzlMath.MOD
            ans %= xzlMath.MOD

    print(ans)
    pass

if __name__ == '__main__':
    t = int(input())
    while t > 0:
        t -= 1
        sol()

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

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

Think step by step.

Response:
```plaintext
1
4999
```


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Ultra-Meow
time_limit_ms: 2500
memory_limit_mb: 256
Description: K1o0n gave you an array $a$ of length $n$, consisting of numbers $1, 2, \ldots, n$. Accept it? Of course! But what to do with it? Of course, calculate $\text{MEOW}(a)$.

Let $\text{MEX}(S, k)$ be the $k$-th positive (strictly greater than zero) integer in ascending order that is not present in the set $S$. Denote $\text{MEOW}(a)$ as the sum of $\text{MEX}(b, |b| + 1)$, over all distinct subsets $b$ of the array $a$.

Examples of $\text{MEX}(S, k)$ values for sets:

  * $\text{MEX}(\\{3,2\\}, 1) = 1$, because $1$ is the first positive integer not present in the set;    * $\text{MEX}(\\{4,2,1\\}, 2) = 5$, because the first two positive integers not present in the set are $3$ and $5$;    * $\text{MEX}(\\{\\}, 4) = 4$, because there are no numbers in the empty set, so the first $4$ positive integers not present in it are $1, 2, 3, 4$.

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

In a single line of each test case, an integer $n$ ($1 \le n \le 5000$) is entered, the size of the array of gifted numbers.

It is guaranteed that the sum of $n^2$ over all test cases does not exceed $25 \cdot 10^6$.

Output: For each test case, output a single number — $\text{MEOW}(a)$. Since it may be very large, output it modulo $10^9 + 7$.

Examples:
input:
5
2
3
4999
5
1
output:
12
31
354226409
184
4

Note:

Buggy Code:
import java.util.Scanner;

public class G {
    static Scanner in = new Scanner(System.in);
    static long[][] triangle;

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

        triangle = new long[5000][5000];
        for (int i = 0; i < 5000; i++) {
            triangle[i][0] = 1;
        }
        for (int i = 1; i < 5000; i++) {
            for (int j = 1; j <= i; j++) {
                triangle[i][j] = triangle[i - 1][j - 1] + triangle[i - 1][j];
                triangle[i][j] %= 1000000007;
            }
        }

        while (t-- > 0) {
            solve();
        }
    }

    public static void solve() {
        int n = in.nextInt();
        if(n==1) {
            System.out.println(4);
            return;
        }
        long ans = 1;
        int left = 1;
        int right = n-2;
        int e = 2*n + 1;
        for (int i = 2; i < n ; i++) {
            int x = 0;
            int y = i-1;
            if(y > right) {
                int temp=y;
                if(y%2 != right%2) {
                    y=right-1;
                }
                else y=right;
                x += (temp-y)/2;
            }
            long temp =com(right , y)*com(left , x);
            temp %= 1000000007;
            ans += i*temp;
            ans %= 1000000007;
            while (y >= 2) {
                y-=2;
                x++;
                temp =com(right , y)*com(left , x);
                temp %= 1000000007;
                ans += i*temp;
                ans %= 1000000007;
            }
            left++;
            right--;
        }
        int s=0;
        if(n%2 == 0) s=n+1;
        else {
            ans += n*com(n-1,n/2);
            ans %= 1000000007;
            s=n+2;
        }
        for (int i = s; i <= e; i+=2) {
            ans += i*com(n , i/2);
            ans %= 1000000007;
        }
        System.out.println(ans);
    }

    public static long com(int n , int r) {
        return triangle[n][r];
    }

    public static void printArray(int[] arr) {
        StringBuilder sb = new StringBuilder();
        for (int x : arr) {
            sb.append(x).append(" ");
        }
        System.out.println(sb);
    }

    public static int[] readArray(int n) {
        int[] arr = new int[n];
        for (int i = 0; i < n; i++) {
            arr[i] = in.nextInt();
        }
        return arr;
    }

    public static int[] freq(String s) {
        int[] arr = new int[26];
        for (char c : s.toCharArray()) {
            arr[c - 'a']++;
        }
        return arr;
    }

    public static long gcd(long a, long b) {
        if (a > b)
            a = (a + b) - (b = a);
        if (a == 0L)
            return b;
        return gcd(b % a, a);
    }

    public static boolean isPrime(long n) {
        if (n < 2) return false;
        if (n == 2 || n == 3) return true;
        if (n % 2 == 0 || n % 3 == 0) return false;
        long sqrtN = (long) Math.sqrt(n) + 1;
        for (long i = 6L; i <= sqrtN; i += 6) {
            if (n % (i - 1) == 0 || n % (i + 1) == 0) return false;
        }
        return true;
    }
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases and scenarios that maximize the failure probability. The code appears to have issues with combinatorial calculations and the handling of specific edge cases. Here are the steps to create a small-scale test input:

1. **Edge Case for Small `n`**:
   - Test with `n = 1` to check if the special handling for `n = 1` works correctly.
   - Test with `n = 2` and `n = 3` to verify the basic combinatorial logic.

2. **Edge Case for Odd and Even `n`**:
   - Test with an odd `n` to ensure the handling of the middle element is correct.
   - Test with an even `n` to check the handling of the middle elements.

3. **Boundary Case for Large `n`**:
   - Test with a large `n` within the constraints (e.g., `n = 4999`) to ensure the combinatorial calculations do not overflow or produce incorrect results.

4. **Multiple Test Cases**:
   - Include multiple test cases to ensure the code handles different scenarios correctly.

Here is the test input that meets these criteria:

```plaintext
5
1
2
3
4999
4
```

This test input includes:
- `n = 1` to check the special handling.
- `n = 2` and `n = 3` to verify basic combinatorial logic.
- `n = 4999` to test the boundary case for large `n`.
- `n = 4` to test an even `n`.

This should help expose any bugs in the combinatorial calculations and edge case handling in the provided code.


**Task:**
Generate a challenging test input that exposes the bug in the buggy code of the algorithm problem:
Algorithm Problem:
Title: Kosuke's Sloth
time_limit_ms: 1000
memory_limit_mb: 256
Description: Kosuke is too lazy. He will not give you any legend, just the task:

Fibonacci numbers are defined as follows:

  * $f(1)=f(2)=1$.    * $f(n)=f(n-1)+f(n-2)$ $(3\le n)$

We denote $G(n,k)$ as an index of the $n$-th Fibonacci number that is divisible by $k$. For given $n$ and $k$, compute $G(n,k)$.

As this number can be too big, output it by modulo $10^9+7$.

For example: $G(3,2)=9$ because the $3$-rd Fibonacci number that is divisible by $2$ is $34$. $[1,1,\textbf{2},3,5,\textbf{8},13,21,\textbf{34}]$.

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

The first and only line contains two integers $n$ and $k$ ($1 \le n \le 10^{18}$, $1 \le k \le 10^5$).

It is guaranteed that the sum of $k$ across all test cases does not exceed $10^6$.

Output: For each test case, output the only number: the value $G(n,k)$ taken by modulo $10^9+7$.

Examples:
input:
3
3 2
100 1
1000000000000 1377
output:
9
100
999244007

Note:

Buggy Code:
#define _USE_MATH_DEFINES
#include<algorithm>
#include<vector>
#include<numeric>
#include<cmath>
#include<fstream>
#include<random>
#include<set>
#include<unordered_set>
#include<string>
#include<map>
#include<unordered_map>
#include<queue>
#include<stack>
#include<fstream>
#include<time.h>
using namespace std;

#define all(a) a.begin(), a.end()
#define Max(a) *max_element(all(a))
#define Min(a) *min_element(all(a))
#define FOR(a, c) for (int(a) = 0; (a) < (c); (a)++)
#define int long long

#define ld long double
#define MP make_pair
#define pb push_back
#include <iostream>

int modd=1e9+7;
int powe(int a,int n){
int ans=1;
while(n>0){
if(n%2){
n--;
ans*=a;
ans%=modd;
}else{
n/=2;
a*=a;
a%=modd;
}
}

return ans;
}

int factorial[200005];
int modInverse(int a, int p) {
int m0 = p, t, q;
int x0 = 0, x1 = 1;
if (p == 1)
return 0;

while (a > 1) {
q = a / p;
t = p;
p = a % p;
a = t;
t = x0;
x0 = x1 - q * x0;
x1 = t;
}

if (x1 < 0)
x1 += m0;
return x1;
}

int combi(int n,int r){
if(n<r){
return 0;
}
return (factorial[n]*((modInverse(factorial[n-r],modd)*modInverse(factorial[r],modd))%modd))%modd;
}

int gcdd(int a,int b){
if(b>a){
return gcdd(b,a);
}
if(b==0){
return a;
}
return gcdd(b,a%b);
}

int findGCD(vector<int> &v){

int n=v.size();
int g=abs(v[0]);
for(int i=0;i<n;i++){
g=gcdd(g,v[i]);
}
return g;
}
void solve(){

  int n,k; cin>>n>>k;
  vector<int>fib(3e5,1);
 // cout<<fib[0]<<" "<<fib[1]<<" ";
 if(k==1){
  cout<<n<<endl;
  return;
 }
  for(int i=2;i<3e5;i++){
    fib[i]=fib[i-1]+fib[i-2];
    fib[i]%=k;
    if(fib[i]==0){
      int sum=0;
      for(int j=0;j<=i;j++){
        sum+=n;
        sum%=modd;
      }
      cout<<sum<<endl;
      return;
    }
  }
 // cout<<endl;

}

signed main(){

ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
 int t;
cin>>t;
while(t--){

solve();
}
  return 0;
}

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

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

Think step by step.

Response:
To generate a challenging test input that exposes the bug in the provided code, we need to focus on edge cases or scenarios that can cause the code to fail. The key points to consider are:

1. **Large Values of \( n \)**: The code should handle very large values of \( n \) (up to \( 10^{18} \)) efficiently.
2. **Small Values of \( k \)**: The code should correctly handle small values of \( k \) (e.g., \( k = 1 \)).
3. **Large Values of \( k \)**: The code should correctly handle large values of \( k \) (up to \( 10^5 \)).
4. **Edge Cases for \( k \)**: Values of \( k \) that are not divisors of Fibonacci numbers in a small range should be tested.

Given the code's structure, it seems to have issues with handling large values of \( n \) and \( k \). Specifically, the code uses a fixed size array for the Fibonacci sequence and does not handle the modulo operation efficiently.

Let's generate a test input that includes these edge cases:

1. A large value of \( n \) with a small value of \( k \).
2. A large value of \( n \) with a large value of \( k \).
3. A small value of \( n \) with a large value of \( k \).

Here is the test input:

```plaintext
3
1000000000000 1
1000000000000 100000
10 99999
```

This test input includes:
1. A large \( n \) with \( k = 1 \) (should output \( n \)).
2. A large \( n \) with a large \( k \) (to test the efficiency and correctness of the algorithm).
3. A small \( n \) with a large \( k \) (to test the edge case where \( k \) is not a divisor of any Fibonacci number in a small range).

This should help expose any bugs in the code.