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;
  template <typename Arg1>
  void __f(const char* name, Arg1&& arg1) {
    cerr << name << " : " << arg1 << endl;
  }
  template <typename Arg1, typename... Args>
  void __f(const char* names, Arg1&& arg1, Args&&... args) {
    const char* comma = strchr(names + 1, ',');
    cerr.write(names, comma - names) << " : " << arg1 << " | ";
    __f(comma + 1, args...);
  }
  long long modI(long long a, long long m);
  long long gcd(long long a, long long b);
  long long powM(long long x, long long y, long long m);
  void py(int ok) {
    if (ok == 1) {
      cout << "YES"
           << "\n";
    } else {
      cout << "NO"
           << "\n";
    }
  }
  void swap(long long& a, long long& b) {
    long long tp = a;
    a = b;
    b = tp;
  }
  long long gcd(long long x, long long y) {
    if (x == 0) return y;
    return gcd(y % x, x);
  }
  long long lcm(long long x, long long y) {
    if (x == 0 || y == 0) {
      cout << "Not valid lcm"
           << "\n";
    }
    long long LCM = x / gcd(x, y);
    LCM = LCM * y;
    return LCM;
  }
  long long powM(long long x, long long y, long long m) {
    long long ans = 1, r = 1;
    x %= m;
    while (r > 0 && r <= y) {
      if (r & y) {
        ans *= x;
        ans %= m;
      }
      r <<= 1;
      x *= x;
      x %= m;
    }
    return ans;
  }
  long long modI(long long a, long long m) {
    long long m0 = m, y = 0, x = 1;
    if (m == 1) return 0;
    while (a > 1) {
      long long q = a / m;
      long long t = m;
      m = a % m;
      a = t;
      t = y;
      y = x - q * y;
      x = t;
    }
    if (x < 0) x += m0;
    return x;
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(0);
    int tt = 1;
    cin >> tt;
    for (int testcase_no = 1; testcase_no <= tt; testcase_no++) {
      long long n;
      cin >> n;
      long long ans = n;
      for (int i = 1; i < 2 * n; i++) {
        ans *= i;
        ans %= 1000000007;
      }
      cout << ans << "\n";
    }
    return 0;
  }
  