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;
  const long long MOD = 1e9 + 7;
  const long long inf = 1e18 + 7;
  const long long N = 3e5 + 5;
  vector<long long> g[N];
  void setIO(string second) {
    freopen((second + ".in").c_str(), "r", stdin);
    freopen((second + ".out").c_str(), "w", stdout);
  }
  struct mi {
    long long v;
    explicit operator long long() const { return v; }
    mi() { v = 0; }
    mi(long long _v) : v(_v % MOD) { v += (v < 0) * MOD; }
  };
  mi &operator+=(mi &a, mi b) {
    if ((a.v += b.v) >= MOD) a.v -= MOD;
    return a;
  }
  mi &operator-=(mi &a, mi b) {
    if ((a.v -= b.v) < 0) a.v += MOD;
    return a;
  }
  mi operator+(mi a, mi b) { return a += b; }
  mi operator-(mi a, mi b) { return a -= b; }
  mi operator*(mi a, mi b) { return mi((long long)a.v * b.v); }
  mi &operator*=(mi &a, mi b) { return a = a * b; }
  mi pow(mi a, long long p) {
    assert(p >= 0);
    return p == 0 ? 1 : pow(a * a, p / 2) * (p & 1 ? a : 1);
  }
  mi inv(mi a) {
    assert(a.v != 0);
    return pow(a, MOD - 2);
  }
  mi operator/(mi a, mi b) { return a * inv(b); }
  template <typename T_vector>
  void output_vector(const T_vector &v, bool add_one = false,
                     long long start = -1, long long end = -1) {
    if (start < 0) start = 0;
    if (end < 0) end = v.size();
    for (long long i = start; i < end; i++)
      cout << v[i] + (add_one ? 1 : 0) << (i < end - 1 ? ' ' : '\n');
  }
  void yes() {
    cout << "YES"
         << "\n";
    ;
  }
  void no() {
    cout << "NO"
         << "\n";
    ;
  }
  mi fact[N];
  void pre() {
    fact[1] = (mi)1;
    fact[2] = (mi)1;
    for (long long i = 3; i < N; i++) {
      fact[i] = (fact[i - 1] * i);
    }
  }
  void solve_test_case() {
    long long n;
    cin >> n;
    cout << (long long)fact[2 * n] << "\n";
    ;
  }
  signed main() {
    long long t;
    cin >> t;
    pre();
    for (long long i = 1; i < t + 1; i++) {
      solve_test_case();
    }
    return 0;
  }
  