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>
  #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;
  }
  