#include <bits/stdc++.h>
using namespace std;
int powmod(int x, int y, int p) {
  int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y % 2 == 1) res = (res * x) % p;
    y /= 2;
    x = (x * x) % p;
  }
  return res;
}
string dectobin(int x) {
  string s = "";
  while (x > 0) {
    int t = x % 2;
    s.push_back(t + '0');
    x /= 2;
  }
  reverse(s.begin(), s.end());
  if (s.compare("") == 0)
    return "0";
  else
    return s;
}
int long long bintodec(string s) {
  int long long ans = 0;
  int long long n = s.size();
  for (int long long i = n - 1; i >= 0; i--) {
    if (s[i] == '1') ans += pow(2, n - i - 1);
  }
  return ans;
}
int find(int k, int n) { return ((n & (1 << (k - 1))) >> (k - 1)); }
int long long f[100000 + 1];
int32_t main() {
  f[0] = 1;
  f[1] = 1;
  for (int i = 2; i <= 100000; i++) {
    f[i] = ((i << 1) - 1ll) * f[i - 1] % 1000000007 * (i << 1) % 1000000007;
  }
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int long long n;
    cin >> n;
    cout << f[n] << "\n";
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
int powmod(int x, int y, int p) {
  int res = 1;
  x = x % p;
  if (x == 0) return 0;
  while (y > 0) {
    if (y % 2 == 1) res = (res * x) % p;
    y /= 2;
    x = (x * x) % p;
  }
  return res;
}
string dectobin(int x) {
  string s = "";
  while (x > 0) {
    int t = x % 2;
    s.push_back(t + '0');
    x /= 2;
  }
  reverse(s.begin(), s.end());
  if (s.compare("") == 0)
    return "0";
  else
    return s;
}
int long long bintodec(string s) {
  int long long ans = 0;
  int long long n = s.size();
  for (int long long i = n - 1; i >= 0; i--) {
    if (s[i] == '1') ans += pow(2, n - i - 1);
  }
  return ans;
}
int find(int k, int n) { return ((n & (1 << (k - 1))) >> (k - 1)); }
int long long f[100000 + 1];
int32_t main() {
  f[0] = 1;
  f[1] = 1;
  for (int i = 2; i <= 100000; i++) {
    f[i] = ((i << 1) - 1ll) * f[i - 1] % 1000000007 * (i << 1) % 1000000007;
  }
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    int long long n;
    cin >> n;
    cout << f[n] << "\n";
  }
}

```