#include <bits/stdc++.h>
using namespace std;
const long long N = 2e6 + 6;
const long long P = 1e9 + 7;
long long read() {
  long long an = 0, ff = 1;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') ff = -1;
  for (; c >= '0' && c <= '9'; c = getchar())
    an = (an << 3) + (an << 1) + c - '0';
  return an * ff;
}
long long Add(long long a, long long b) {
  return a + b >= P ? a + b - P : a + b;
}
long long Sub(long long a, long long b) {
  return a - b < 0 ? a - b + P : a - b;
}
long long Mul(long long a, long long b) {
  long long res = 1ll * a * b;
  return res >= P ? res % P : res;
}
long long Pow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = Mul(res, a);
    a = Mul(a, a);
    b >>= 1;
  }
  return res;
}
long long Inv(long long a) { return Pow(a, P - 2); }
long long f[N];
signed main() {
  f[1] = f[0] = 1;
  for (long long i = 1; i <= 2e5; i++) f[i] = f[i - 1] * i % P;
  long long T = read();
  while (T--) {
    long long x = read();
    x *= 2;
    if (f[x - 1] * x < P)
      cout << f[x] / 2 << endl;
    else
      cout << f[x] * Inv(2) % P << endl;
  }
  return 0;
}

### Prompt

Create a solution in Cpp for 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>
using namespace std;
const long long N = 2e6 + 6;
const long long P = 1e9 + 7;
long long read() {
  long long an = 0, ff = 1;
  char c = getchar();
  for (; c < '0' || c > '9'; c = getchar())
    if (c == '-') ff = -1;
  for (; c >= '0' && c <= '9'; c = getchar())
    an = (an << 3) + (an << 1) + c - '0';
  return an * ff;
}
long long Add(long long a, long long b) {
  return a + b >= P ? a + b - P : a + b;
}
long long Sub(long long a, long long b) {
  return a - b < 0 ? a - b + P : a - b;
}
long long Mul(long long a, long long b) {
  long long res = 1ll * a * b;
  return res >= P ? res % P : res;
}
long long Pow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = Mul(res, a);
    a = Mul(a, a);
    b >>= 1;
  }
  return res;
}
long long Inv(long long a) { return Pow(a, P - 2); }
long long f[N];
signed main() {
  f[1] = f[0] = 1;
  for (long long i = 1; i <= 2e5; i++) f[i] = f[i - 1] * i % P;
  long long T = read();
  while (T--) {
    long long x = read();
    x *= 2;
    if (f[x - 1] * x < P)
      cout << f[x] / 2 << endl;
    else
      cout << f[x] * Inv(2) % P << endl;
  }
  return 0;
}

```