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: import java.io.*;
  import java.util.*;
  
  public class A {	
  		
  	static long mod=(long)1e9+7;
  	
  	public static void main(String[] args) throws Exception
  	{
  		FastReader f = new FastReader();
  		StringBuffer sb=new StringBuffer();
  		
  		int test=f.nextInt();
  		out:
  		while(test-->0)
  		{
  			long n=2*f.nextLong();
  			long ans=1;
  			for(int i=3;i<=n;i++)
  				ans=(ans*i)%mod;
  			sb.append(ans+"\n");
  		}
  		System.out.println(sb);
  	}
  	
  	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; 
  	    } 
  	} 
  }