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 double PI = 3.141592653589793;
  const int Mod[] = {998244353, (int)1e9 + 7, 1223456789, 1234567891, 2147483647};
  const int mod = Mod[1];
  const int N = 2e6 + 5;
  const int NN = 1e7 + 5;
  const int M = 1e3 + 5;
  const int MM = 1e4;
  const long long oo = 9223372036854775807ll;
  typedef bool (*func)(int &, int);
  typedef bool (*funcl)(long long &, long long);
  template <class t>
  bool add(t &x, t y) {
    x += y;
    if (x >= mod) {
      x -= mod;
      return 0;
    }
    return 1;
  }
  template <class t>
  bool mius(t &x, t y) {
    x -= y;
    if (x < 0) {
      x += mod;
      return 0;
    }
    return 1;
  }
  template <class t>
  bool mini(t &x, t y) {
    if (x > y) {
      x = y;
      return 1;
    }
    return 0;
  }
  template <class t>
  bool maxi(t &x, t y) {
    if (x < y) {
      x = y;
      return 1;
    }
    return 0;
  }
  template <class t>
  bool doxor(t &x, t y) {
    x ^= y;
    return (x > 0);
  }
  template <class T>
  class DSU {
   private:
    T n;
    vector<T> cha;
  
   public:
    DSU(const T &n = 0) : n(n + 7), cha(n + 7, -1) {}
    T root(T u) { return (cha[u] < 0 ? u : cha[u] = root(cha[u])); }
    bool join(T u, T v) {
      if ((v = root(v)) == (u = root(u))) return 0;
      if (cha[u] > cha[v]) swap(u, v);
      cha[u] += cha[v];
      cha[v] = u;
      return 1;
    }
  };
  template <class T, class TT, class up>
  class FenWick {
   private:
    T n;
    TT init;
    vector<TT> bit;
    up update;
  
   public:
    FenWick(const T &n = 0, const up &update = nullptr, const TT &init = 0)
        : n(n + 7), init(init), bit(n + 7, init), update(update) {}
    void adj(T x, TT val) {
      for (; x <= n; x += x & -x) update(bit[x], val);
    }
    TT get(T x) {
      TT res = init;
      for (; x > 0; x -= x & -x) update(res, bit[x]);
      return res;
    }
  };
  void file(string name = "") {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    return;
  }
  long long gt[N], rgt[N];
  long long power(long long a, long long b) {
    if (b == 0) return 1ll;
    if (b == 1) return a;
    long long tam = 1ll * power(a, b / 2);
    tam = 1ll * tam * tam % mod;
    if (b & 1) return 1ll * tam * a % mod;
    return tam;
  }
  void prepare() {
    gt[0] = 1;
    for (int i = 1; i <= N; i++) gt[i] = 1ll * gt[i - 1] * i % mod;
  }
  int nCk(int n, int k) {
    return 1ll * gt[n] * power(gt[n - k], mod - 2) % mod * power(gt[k], mod - 2) %
           mod;
  }
  void solve(int test = -1) {
    int n;
    cin >> n;
    long long res = 1;
    for (int i = 1; i <= n; i++) {
      res *= n * 2 - i;
      res %= mod;
    }
    res *= gt[n];
    res %= mod;
    cout << res;
    return;
  }
  int32_t main() {
    file();
    int T = 1;
    prepare();
    cin >> T;
    for (int i = 1; i <= T; i++) {
      solve(i);
      cout << '\n';
    }
    return 0;
  }
  