#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long int expo(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  return res;
}
long long int mmiprime(long long int a, long long int b) {
  return expo(a, b - 2, b);
}
void extendgcd(long long int a, long long int b, long long int* v) {
  if (b == 0) {
    v[0] = 1;
    v[1] = 0;
    v[2] = a;
    return;
  }
  extendgcd(b, a % b, v);
  long long int x = v[1];
  v[1] = v[0] - v[1] * (a / b);
  v[0] = x;
  return;
}
long long int mod_add(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
long long int mod_mul(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
long long int mod_sub(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
long long int mod_div(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (mod_mul(a, mmiprime(b, m), m) + m) % m;
}
vector<long long int> sieve(int n) {
  int* arr = new int[n + 1]();
  vector<long long int> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
bool sortbysec(const pair<int, int>& a, const pair<int, int>& b) {
  return (a.second < b.second);
}
const long long INF = 1e18;
int factorial[3000001];
int invfac[3000001];
int v[2];
void fac() {
  factorial[0] = 1;
  invfac[0] = 1;
  for (int i = 1; i < 3000001; i++) {
    factorial[i] = mod_mul(factorial[i - 1], i, 1000000007);
  }
  invfac[3000000] = mmiprime(factorial[3000000], 1000000007);
  for (int i = 3000000 - 1; i > 0; i--) {
    invfac[i] = mod_mul(invfac[i + 1], i + 1, 1000000007);
  }
}
int ncr(int a, int b) {
  if (b > a) return 0;
  int x = invfac[b];
  int y = invfac[a - b];
  int ans = mod_mul(factorial[a], x, 1000000007);
  ans = mod_mul(ans, y, 1000000007);
  return ans;
}
void sk_kr() {
  int n;
  cin >> n;
  n *= 2;
  long long int ans =
      mod_mul(factorial[n], mmiprime(2, 1000000007), 1000000007);
  cout << ans << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  auto begin = chrono::high_resolution_clock::now();
  int t;
  fac();
  t = 1;
  cin >> t;
  while (t--) {
    sk_kr();
  }
  auto end = chrono::high_resolution_clock::now();
  auto duration = chrono::duration_cast<chrono::milliseconds>(end - begin);
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

CQXYM is counting permutations length of 2n.

A permutation is an array consisting of n distinct integers from 1 to n in arbitrary order. For example, [2,3,1,5,4] is a permutation, but [1,2,2] is not a permutation (2 appears twice in the array) and [1,3,4] is also not a permutation (n=3 but there is 4 in the array).

A permutation p(length of 2n) will be counted only if the number of i satisfying p_i<p_{i+1} is no less than n. For example:

  * Permutation [1, 2, 3, 4] will count, because the number of such i that p_i<p_{i+1} equals 3 (i = 1, i = 2, i = 3).
  * Permutation [3, 2, 1, 4] won't count, because the number of such i that p_i<p_{i+1} equals 1 (i = 3). 



CQXYM wants you to help him to count the number of such permutations modulo 1000000007 (10^9+7).

In addition, [modulo operation](https://en.wikipedia.org/wiki/Modulo_operation) is to get the remainder. For example:

  * 7 mod 3=1, because 7 = 3 ⋅ 2 + 1, 
  * 15 mod 4=3, because 15 = 4 ⋅ 3 + 3. 

Input

The input consists of multiple test cases. 

The first line contains an integer t (t ≥ 1) — the number of test cases. The description of the test cases follows.

Only one line of each test case contains an integer n(1 ≤ n ≤ 10^5).

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

Output

For each test case, print the answer in a single line.

Example

Input


4
1
2
9
91234


Output


1
12
830455698
890287984

Note

n=1, there is only one permutation that satisfies the condition: [1,2].

In permutation [1,2], p_1<p_2, and there is one i=1 satisfy the condition. Since 1 ≥ n, this permutation should be counted. In permutation [2,1], p_1>p_2. Because 0<n, this permutation should not be counted.

n=2, there are 12 permutations: [1,2,3,4],[1,2,4,3],[1,3,2,4],[1,3,4,2],[1,4,2,3],[2,1,3,4],[2,3,1,4],[2,3,4,1],[2,4,1,3],[3,1,2,4],[3,4,1,2],[4,1,2,3].

### Response

```cpp
#include <bits/stdc++.h>
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
using namespace std;
long long int expo(long long int a, long long int b, long long int mod) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b = b >> 1;
  }
  return res;
}
long long int mmiprime(long long int a, long long int b) {
  return expo(a, b - 2, b);
}
void extendgcd(long long int a, long long int b, long long int* v) {
  if (b == 0) {
    v[0] = 1;
    v[1] = 0;
    v[2] = a;
    return;
  }
  extendgcd(b, a % b, v);
  long long int x = v[1];
  v[1] = v[0] - v[1] * (a / b);
  v[0] = x;
  return;
}
long long int mod_add(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a + b) % m) + m) % m;
}
long long int mod_mul(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a * b) % m) + m) % m;
}
long long int mod_sub(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (((a - b) % m) + m) % m;
}
long long int mod_div(long long int a, long long int b, long long int m) {
  a = a % m;
  b = b % m;
  return (mod_mul(a, mmiprime(b, m), m) + m) % m;
}
vector<long long int> sieve(int n) {
  int* arr = new int[n + 1]();
  vector<long long int> vect;
  for (int i = 2; i <= n; i++)
    if (arr[i] == 0) {
      vect.push_back(i);
      for (int j = 2 * i; j <= n; j += i) arr[j] = 1;
    }
  return vect;
}
bool sortbysec(const pair<int, int>& a, const pair<int, int>& b) {
  return (a.second < b.second);
}
const long long INF = 1e18;
int factorial[3000001];
int invfac[3000001];
int v[2];
void fac() {
  factorial[0] = 1;
  invfac[0] = 1;
  for (int i = 1; i < 3000001; i++) {
    factorial[i] = mod_mul(factorial[i - 1], i, 1000000007);
  }
  invfac[3000000] = mmiprime(factorial[3000000], 1000000007);
  for (int i = 3000000 - 1; i > 0; i--) {
    invfac[i] = mod_mul(invfac[i + 1], i + 1, 1000000007);
  }
}
int ncr(int a, int b) {
  if (b > a) return 0;
  int x = invfac[b];
  int y = invfac[a - b];
  int ans = mod_mul(factorial[a], x, 1000000007);
  ans = mod_mul(ans, y, 1000000007);
  return ans;
}
void sk_kr() {
  int n;
  cin >> n;
  n *= 2;
  long long int ans =
      mod_mul(factorial[n], mmiprime(2, 1000000007), 1000000007);
  cout << ans << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  auto begin = chrono::high_resolution_clock::now();
  int t;
  fac();
  t = 1;
  cin >> t;
  while (t--) {
    sk_kr();
  }
  auto end = chrono::high_resolution_clock::now();
  auto duration = chrono::duration_cast<chrono::milliseconds>(end - begin);
  return 0;
}

```