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;
  long long Pow(long long b, long long p) {
    return p == 0 ? 1 : b * Pow(b, p - 1);
  }
  long long gcd(long long a, long long b) { return (b == 0) ? a : gcd(b, a % b); }
  long long lcm(long long a, long long b) { return (a / gcd(a, b)) * b; }
  string to_String(long long n) {
    string s;
    while (n) {
      s.push_back(n % 10 + '0');
      n /= 10;
    }
    reverse(s.begin(), s.end());
    return s;
  }
  long long to_int(string s) {
    long long n = 0, i = 0, j = 1;
    for (long long i = 0; i < s.size(); i++) {
      n += (s[s.size() - (i + 1)] - '0') * j;
      j *= 10;
    }
    return n;
  }
  long long BigMod(long long b, long long p, long long m) {
    if (p == 0) return 1;
    long long ans = BigMod(b, p / 2, m);
    ans = (ans * ans) % m;
    if (p % 2 == 1) ans = (ans * b) % m;
    return ans;
  }
  long long binarySearch(long long arr[], long long l, long long r, long long x) {
    if (r >= l) {
      long long mid = l + (r - l) / 2;
      if (arr[mid] == x) return mid;
      if (arr[mid] > x) return binarySearch(arr, l, mid - 1, x);
      return binarySearch(arr, mid + 1, r, x);
    }
    return -1;
  }
  long long ModInv(long long a, long long m) { return BigMod(a, m - 2, m); }
  long long egcd(long long a, long long b, long long &x, long long &y) {
    if (b == 0) {
      x = 1;
      y = 0;
      return a;
    }
    long long x1, y1;
    long long d = egcd(b, a % b, x1, y1);
    x = y1;
    y = x1 - (a / b) * y1;
    return d;
  }
  long long fact(long long n) {
    if (n == 1)
      return 1;
    else
      return (n * fact(n - 1));
  }
  long long fun(long long n) {}
  void solve() {
    long long n;
    cin >> n;
    long long cnt = 1;
    for (long long i = n * 2; i > 2; i--) {
      cnt = (cnt * i) % 1000000007;
    }
    cnt = cnt % 1000000007;
    cout << cnt << "\n";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ;
    long long t;
    cin >> t;
    while (t--) {
      solve();
    };
    return 0;
  }
  void sieve(vector<long long> &v) {
    v.assign(1000000007, 0);
    v[0] = v[1] = 1;
    for (int i = 4; i <= 1000000007; i += 2) v[i] = 1;
    for (int i = 3; i * i <= 1000000007; i += 2) {
      if (v[i] == 0) {
        for (int j = i * i; j <= 1000000007; j += i * 2) v[j] = 1;
      }
    }
  }
  