description: 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].
solution: #include <bits/stdc++.h>
  using namespace std;
  using ll = long long;
  using vb = vector<bool>;
  using vvb = vector<vb>;
  using vi = vector<int>;
  using vvi = vector<vi>;
  using vl = vector<ll>;
  using vvl = vector<vl>;
  using vc = vector<char>;
  using vvc = vector<vc>;
  using vs = vector<string>;
  const ll mod = 1e9 + 7, inf = 1e18;
  void setIO() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
  }
  int main() {
    setIO();
    int t;
    cin >> t;
    while (t--) {
      ll n;
      cin >> n;
      ll m = 1;
      for (int i = 3; i <= 2 * n; i++) m = m * i % 1000000007;
      cout << m << endl;
    }
    return 0;
  }
  