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.util.*;

public class A {
	static class InputReader {
		BufferedReader reader;
		StringTokenizer tokenizer;
		public InputReader(InputStream stream) {
			reader = new BufferedReader(new InputStreamReader(stream), 32768);
			tokenizer = null;
		}
		String next() {
			while (tokenizer == null || !tokenizer.hasMoreTokens()) {
				try {
					tokenizer = new StringTokenizer(reader.readLine());
				} catch (IOException e) {
					throw new RuntimeException(e);
				}
			}
			return tokenizer.nextToken();
		}
		public int ni() {
			return Integer.parseInt(next());
		}
		public long nl() {
			return Long.parseLong(next());
		}
		public double nd() {
			return Double.parseDouble(next());
		}
		
		public int[] ra(int n ) {
			int[] a = new int[n];
			for (int i = 0; i < n; i++) {
				a[i] = r.ni();
			}
			return a;
		}
	}
	
	public static void sort(int[] a) {
		List<Integer> l = new ArrayList<>();
		for (int i: a) {
			l.add(i);
		}
		Collections.sort(l);
		for (int i = 0; i < a.length; i++) {
			a[i] = l.get(i);
		}
	}
	
	public static void sortL(long[] a) {
		List<Long> l = new ArrayList<>();
		for (long i: a) {
			l.add(i);
		}
		Collections.sort(l);
		for (int i = 0; i < a.length; i++) {
			a[i] = l.get(i);
		}
	}
	
	static InputReader r = new InputReader(System.in);
	static PrintWriter pw = new PrintWriter(System.out);
	static int m = 1000000007;
	
	
	static void solve() {
		int n = r.ni();
		
		long res = 1;
		
		for(int i = 3; i <= n * 2; i++) {
			res = (res * i) % m;
		}
		
		pw.println(res);
	}
	
	public static void main(String[] args) {
		int TT = r.ni();
		
		while (TT-- > 0) {
			solve();
		}
		
		pw.close();
	}
}
