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;
  void printlist(vector<long long> temp) {
    for (long long i = 0; i < temp.size(); ++i) {
      cout << temp[i] << " ";
    }
    cout << "\n";
  }
  long long mod = 1e9 + 7;
  struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
      x += 0x9e3779b97f4a7c15;
      x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
      x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
      return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const {
      static const uint64_t FIXED_RANDOM =
          chrono::steady_clock::now().time_since_epoch().count();
      return splitmix64(x + FIXED_RANDOM);
    }
  };
  long long gcd(long long a, long long b) {
    if (b == 0) return a;
    return gcd(b, a % b);
  }
  long long mul(long long a, long long b) {
    return ((a % mod) * (b % mod)) % mod;
  }
  long long binexp(long long a, long long b) {
    if (b == 0) return 1;
    long long half = binexp(a, b / 2);
    if ((b % 2) == 0)
      return mul(half, half);
    else
      return mul(a, mul(half, half));
  }
  long long mod_inv(long long n) { return binexp(n, mod - 2); }
  long long msb(long long n) {
    n |= (n >> 1);
    n |= (n >> 2);
    n |= (n >> 2);
    n |= (n >> 8);
    n |= (n >> 16);
    n |= (n >> 32);
    n = n + 1;
    return (n >> 1);
  }
  long long solve(long long n, long long a, long long low, long long high,
                  long long m) {
    long long cn1 = 0;
    long long cn2 = 0;
    long long l = low;
    long long r = high;
    while (low <= high) {
      long long mid = low + (high - low) / 2;
      if ((n * a + mid - m) <= r && (n * a + mid - m) >= l) {
        return mid;
      } else if ((n * a + mid - m) > r) {
        cn1++;
        high = mid - 1;
      } else if ((n * a + mid - m) < l) {
        cn2++;
        low = mid + 1;
      }
    }
    if (cn1 <= cn2) return -1;
    return -2;
  }
  void precompute(long long fact[]) {
    fact[0] = 1;
    for (long long i = 1; i < 200005; i++) {
      fact[i] = (fact[i - 1] * i) % mod;
    }
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(NULL);
    long long fact[200005];
    precompute(fact);
    int t;
    cin >> t;
    while (t--) {
      long long n;
      cin >> n;
      cout << (fact[2 * n] * mod_inv(fact[2])) % mod << "\n";
    }
    return 0;
  }
  