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.io.*;
import java.math.BigInteger;
import java.util.*;
public class b729 {

	public static void main(String[] args) throws IOException {
FastReader sc = new FastReader();
BufferedWriter out = new BufferedWriter(
		            new OutputStreamWriter(System.out));
int m = 1000000007;
long [] fac = new long[200006];
fac[0] = 0;
fac[1] = 1;
for(int i = 2 ; i<200003;i++) {
	fac[i] = (fac[i-1] * i)%m;
}

int t = sc.nextInt();
	
for(int o = 0 ; o<t;o++) {
int n = sc.nextInt();
int x = modInverse(2, m);
System.out.println(fac[2*n]*x%m);

}

}
	static int modInverse(int a, int m)
    {
        int m0 = m;
        int y = 0, x = 1;
 
        if (m == 1)
            return 0;
 
        while (a > 1) {
            // q is quotient
            int q = a / m;
 
            int t = m;
 
            // m is remainder now, process
            // same as Euclid's algo
            m = a % m;
            a = t;
            t = y;
 
            // Update x and y
            y = x - q * y;
            x = t;
        }
 
        // Make x positive
        if (x < 0)
            x += m0;
 
        return x;
    }
	}

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;
    }
}
