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 optimize("O2")
  #pragma GCC optimize("O3")
  #pragma GCC optimize("Os")
  #pragma GCC optimize("Ofast")
  #pragma GCC optimize("Ofast")
  #pragma GCC target("avx,avx2,fma")
  using namespace std;
  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));
  }
  const int64_t INF = 1e17;
  bool allsemchar(string s) {
    return (s.find_first_not_of(s[0]) == string::npos);
  }
  long long int fuck(long long int n) {
    return (n == 1 || n == 0) ? 1 : n * fuck(n - 1);
  }
  double value10(double d) { return log10(d); }
  long long int combi(long long int n, long long int k) {
    long long int ans = 1;
    k = k > n - k ? n - k : k;
    long long int j = 1;
    for (; j <= k; j++, n--) {
      if (n % j == 0) {
        ans *= (n / j);
      } else if (ans % j == 0) {
        ans = (ans / j * n);
      } else {
        ans = ((ans * n) / j);
      }
    }
    return (ans);
  }
  long long int digit(long long int n) {
    long long int cnt = 0;
    while (n > 0) {
      n = n / 10;
      cnt++;
    }
    return cnt;
  }
  void removeChar(char* s, char c) {
    long long int j, n = strlen(s);
    for (long long int i = j = 0; i < n; i++)
      if (s[i] != c) s[j++] = s[i];
    s[j] = '\0';
  }
  bool areSame(long long int a[], long long int n) {
    unordered_map<long long int, long long int> m;
    for (int i = 0; i < n; i++) m[a[i]]++;
    if (m.size() == 1)
      return true;
    else
      return false;
  }
  bool palindrom(string s) {
    string u = s;
    reverse(u.begin(), u.end());
    if (s == u) {
      return true;
    } else {
      return false;
    }
  }
  long long int A_mod_B(string N, long long int a) {
    long long int len = N.size();
    long long int ans = 0;
    for (int i = 0; i < len; i++) ans = (ans * 10 + (int)N[i] - '0') % a;
    return ans % a;
  }
  vector<bool> SieveofEratosthenes(int n) {
    vector<bool> primes(n + 1, true);
    primes[0] = 0;
    primes[1] = 0;
    for (int i = 2; i * i <= n; i++) {
      if (primes[i]) {
        for (int j = i * i; j <= n; j += i) {
          primes[j] = false;
        }
      }
    }
    return primes;
  }
  long long int power(long long int a, long long int b) {
    long long int ans = 1;
    while (b) {
      if (b & 1) ans *= a;
      b >>= 1;
      a *= a;
      a %= 1000000007;
      ans %= 1000000007;
    }
    return ans;
  }
  long long int gcdofarray(long long int a[], long long int n) {
    long long int result = a[0];
    for (long long int i = 1; i < n; i++) {
      result = gcd(a[i], result);
      if (result == 1) {
        return 1;
      }
    }
    return result;
  }
  void kishor() {
    long long int n;
    cin >> n;
    long long int ans = 1;
    for (int i = 3; i <= 2 * n; i++) {
      ans = (ans * i) % 1000000007;
    }
    cout << ans << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    long long int t;
    cin >> t;
    while (t--) {
      kishor();
    }
  }
  