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: /*==========================================================================
  *   	AUTHOR:    RonWonWon
  *   	CREATED:   30.09.2021 15:27:15        
  /*==========================================================================*/
  import java.io.*;
  import java.util.*;
  
  public class A {
  	
  	public static void main(String[] args) throws IOException {
  		/*
      in.ini();
      PrintWriter out = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
      */
      long startTime = System.nanoTime();
      int t = in.nextInt(), T = 1;  
      while(t-->0) { 
        int n = 2*in.nextInt();
        long f = 1;
        for(int i=2;i<=n;i++) f = (f*i)%mod;
        out.println((f*modInv(2))%mod);  
     		//out.println("Case #"+T+": "+ans); T++;
      }
      /*
      startTime = System.nanoTime() - startTime;
      System.err.println("Time: "+(startTime/1000000));
      */
      out.flush();
  	}
  	
  	static final long mod=1_000_000_007;
  	
  	static long mul(long a, long b) {
  		return a*b%mod;
  	}
  
  	static long fastPow(long base, long exp) {
  		if (exp==0) return 1;
  		long half=fastPow(base, exp/2);
  		if (exp%2==0) return mul(half, half);
  		return mul(half, mul(half, base));
  	}
  	
  	static long modInv(long x) {
  		return fastPow(x, mod-2);
  	}
  	
  	static long fact(int x) {
  		long ans=1;
  		for (int i=2; i<=x; i++) ans=mul(ans, i);
  		return ans;
  	}
  	
  	static long nCk(int n, int k) {
  		return mul(fact(n), mul(modInv(fact(k)), modInv(fact(n-k))));
  	}
  	
  	static FastScanner in = new FastScanner();
    static PrintWriter out = new PrintWriter(System.out);
  	static int oo = Integer.MAX_VALUE;
  	static long ooo = Long.MAX_VALUE;
  	
  	static class FastScanner {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st = new StringTokenizer("");
  		
  		void ini() throws FileNotFoundException {
  			br = new BufferedReader(new FileReader("input.txt"));
  		}
  		
  		String next() {
  			while(!st.hasMoreTokens())
  				try { st = new StringTokenizer(br.readLine()); }
  				catch(IOException e) {}
  			return st.nextToken();
  		}
  		
  		String nextLine(){
  			try{ return br.readLine(); } 
  			catch(IOException e) { } return "";
  		}
  		
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  		
  		int[] readArray(int n) {
  			int a[] = new int[n];
  			for(int i=0;i<n;i++) a[i] = nextInt();
  			return a;
  		}
  		
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  		
  		long[] readArrayL(int n) {
  			long a[] = new long[n];
  			for(int i=0;i<n;i++) a[i] = nextLong();
  			return a;
  		}
  		
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  		
  		double[] readArrayD(int n) {
  			double a[] = new double[n];
  			for(int i=0;i<n;i++) a[i] = nextDouble();
  			return a;
  		}
  	}
  
  	static final Random random = new Random();
  
  	static void ruffleSort(int[] a){
  		int n = a.length;
  		for(int i=0;i<n;i++){
  			int j = random.nextInt(n), temp = a[j];
  			a[j] = a[i]; a[i] = temp;
  		}
  		Arrays.sort(a); 	
  	}
  	
  	static void ruffleSortL(long[] a){
  		int n = a.length;
  		for(int i=0;i<n;i++){
  			int j = random.nextInt(n);
  			long temp = a[j];
  			a[j] = a[i]; a[i] = temp;
  		}
  		Arrays.sort(a); 	
  	}
  }
  