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 int mod = 1e9 + 7;
  long long int Pow(long long int c, long long int d) {
    return d == 0 ? 1 : c * pow(c, d - 1);
  }
  long long int gcd(long long int a, long long int b) {
    return b == 0 ? a : gcd(b, a % b);
  }
  long long int lcm(long long int a, long long int b) {
    return ((a * b) / gcd(a, b));
  }
  long long int egcd(long long int a, long long int b, long long int &x,
                     long long int &y) {
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    }
    long long int x1, y1;
    long long int gcd = egcd(b, a % b, x1, y1);
    x = y1;
    y = x1 - y1 * (a / b);
    return gcd;
  }
  long long int bigMod(long long int a, long long int b) {
    a %= mod;
    long long int res = 1;
    while (b > 0) {
      if (b & 1) res = (res * a) % mod;
      a = (a * a) % mod;
      b >>= 1;
    }
    return res;
  }
  inline long long int MOD(long long int a) { return (a % mod + mod) % mod; }
  inline long long int modAdd(long long int a, long long int b) {
    return MOD(MOD(a) + MOD(b));
  }
  inline long long int modSub(long long int a, long long int b) {
    return MOD(MOD(a) - MOD(b));
  }
  inline long long int modMul(long long int a, long long int b) {
    return MOD(MOD(a) * MOD(b));
  }
  inline long long int modInv(long long int a) { return bigMod(a, mod - 2); }
  inline long long int modDiv(long long int a, long long int b) {
    a = MOD(a);
    b = modInv(b);
    return (a * b) % mod;
  }
  vector<string> v;
  map<char, long long int> mp;
  bool compare(pair<long long int, pair<long long int, long long int>> p1,
               pair<long long int, pair<long long int, long long int>> p2) {
    if (p1.first == p2.first and p1.second.second > p2.second.second)
      return false;
    else if (p1.first < p2.first)
      return false;
    else
      return true;
  }
  set<string> vs;
  void permute(string str, string out) {
    if (str.size() == 0) {
      vs.insert(out);
      return;
    }
    for (int i = 0; i < str.size(); i++) {
      permute(str.substr(1), out + str[0]);
      rotate(str.begin(), str.begin() + 1, str.end());
    }
  }
  bool isPrime(long long int n) {
    for (long long int i = 2; i * i <= n; i++) {
      if (n % i == 0) return false;
    }
    return true;
  }
  long long int ncr(long long int nn, long long int rr) {
    if (rr > nn) return 0;
    if (rr == 0) return 1;
    if (rr == 1) return nn;
    if (nn == rr) return 1;
    return ncr(nn - 1, rr - 1) + ncr(nn - 1, rr);
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    long long int t;
    cin >> t;
    long long int fact[200001];
    fact[0] = 1;
    for (long long int i = 1; i <= 200000; i++) fact[i] = modMul(fact[i - 1], i);
    while (t--) {
      long long int n;
      cin >> n;
      n = n * 2;
      long long int ans = modDiv(fact[n], 2);
      cout << ans << endl;
    }
    return 0;
  }
  