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].
import java.util.*;
import java.io.*;
public class Main{
	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 int dfs(int node,HashMap<Integer,List<Integer>> map) {
		int ans = 0;
		for(int next : map.get(node)) {
			ans = Math.max(ans,dfs(next,map));
		}
		return ans+1;

	}
	public static long fac(long num)  {
		long ans = 1;
		int mod = (int)1e9+7;
		for(long i = 2;i<=num;i++) {
			ans  =  (ans*i)%mod;
		}
		return ans;
	}
	public static void main(String[] args)throws IOException {
		FastReader sc = new FastReader();
		int t = sc.nextInt();
		while(t-- >0) {
			long n = sc.nextLong();
			int mod = (int)1e9+7;
			long ans = fac(n*2 - 1);
			ans = (ans * n )%mod;
			System.out.println(ans);
		}
	}
	public static boolean isPrime(long num) {
		if(num==1) return false;
		if(num<=3) return true;
		if(num%2==0||num%3==0) return false;
		for(long i =5;i*i<=num;i+=6) {
			if(num%i==0) return false;
		}
		return true;
	}
	public static boolean isPrime(int num) {
		if(num==1) return false;
		if(num<=3) return true;
		if(num%2==0||num%3==0) return false;
		for(int i =5;i*i<=num;i+=6) {
			if(num%i==0) return false;
		}
		return true;
	}
	public static int gcd(int a , int b) {
		if (b == 0) return a;
		return gcd(b, a % b);
	}
	public static long gcd(long a , long b) {
		if (b == 0) return a;
		return gcd(b, a % b);
	}
}
 










