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: // Working program with FastReader
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.Scanner;
  import java.util.StringTokenizer;
  import java.util.*;
  
  
  public class Example {
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader()
          {
              br = new BufferedReader(
                      new InputStreamReader(System.in));
          }
  
          String next()
          {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  }
                  catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() { return Integer.parseInt(next()); }
  
          long nextLong() { return Long.parseLong(next()); }
  
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
  
          String nextLine()
          {
              String str = "";
              try {
                  str = br.readLine();
              }
              catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  
      public static void main(String[] args)
      {
          FastReader sc = new FastReader();
          int t= sc.nextInt();
          while (t>0){
              t--;
              int n= sc.nextInt();
             long ans=1;
             int max=10^9+7;
              max=(int)Math.pow(10,9)+7;
             long aa=facto(2*n,max);
  aa=(aa)%max;
              System.out.println(aa);
  
  
  
          }
  
      }
  
      private static long facto(int i, int max) {
          long ans=1;
          for(long j=3;j<=i;j++){
           //   System.out.println(ans*j);
              ans=(ans*j)%max;
  
          }
          return ans;
      }
  
      private static int lower(int[] ar, int i, int j, int key) {
          int asn=-1;
          while(i<=j){
              int mid=(i+j)/2;
              if(ar[mid]<=key){
                  asn=mid;
                  i=mid+1;
              }else{
                  j=mid-1;
              }
          }
          return asn;
  
  
      }
  }
  