A sequence of n non-negative integers (n ≥ 2) a_1, a_2, ..., a_n is called good if for all i from 1 to n-1 the following condition holds true: $$$a_1 \: \& \: a_2 \: \& \: ... \: \& \: a_i = a_{i+1} \: \& \: a_{i+2} \: \& \: ... \: \& \: a_n, where \&$$$ denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND).

You are given an array a of size n (n ≥ 2). Find the number of permutations p of numbers ranging from 1 to n, for which the sequence a_{p_1}, a_{p_2}, ... ,a_{p_n} is good. Since this number can be large, output it modulo 10^9+7.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^4), denoting the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the size of the array.

The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9) — the elements of the array.

It is guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5.

Output

Output t lines, where the i-th line contains the number of good permutations in the i-th test case modulo 10^9 + 7.

Example

Input


4
3
1 1 1
5
1 2 3 4 5
5
0 2 0 3 0
4
1 3 5 1


Output


6
0
36
4

Note

In the first test case, since all the numbers are equal, whatever permutation we take, the sequence is good. There are a total of 6 permutations possible with numbers from 1 to 3: [1,2,3], [1,3,2], [2,1,3], [2,3,1], [3,1,2], [3,2,1].

In the second test case, it can be proved that no permutation exists for which the sequence is good.

In the third test case, there are a total of 36 permutations for which the sequence is good. One of them is the permutation [1,5,4,2,3] which results in the sequence s=[0,0,3,2,0]. This is a good sequence because 

  *  s_1 = s_2 \: \& \: s_3 \: \& \: s_4 \: \& \: s_5 = 0, 
  *  s_1 \: \& \: s_2 = s_3 \: \& \: s_4 \: \& \: s_5 = 0, 
  *  s_1 \: \& \: s_2 \: \& \: s_3 = s_4 \: \& \: s_5 = 0, 
  *  s_1 \: \& \: s_2 \: \& \: s_3 \: \& \: s_4 = s_5 = 0. 
import java.io.*;
import java.util.*;
public class Main {
	static long mod=1_000_000_000+7;
	static long[] facts=new long[3_000_00];
	static long mul(long a, long b) {
		return a*b%mod;
	}
	
	static long add(long a, long b) {
		return (a+b)%mod;
	}
	
	static void precomp() {
		facts[0]=1;
		for (int i=1; i<facts.length; i++) facts[i]=mul(facts[i-1], i);
	}
	
	static long exp(long base, long pow) {
		if (pow==0) return 1;
		long half=exp(base, pow/2);
		if (pow%2==0) return mul(half, half);
		return mul(half, mul(half, base));
	}
	
	static long modInv(long x) {
		return exp(x, mod-2);
	}
	
	static long nCk(int n, int k) {
		if(n<k) return 0;
		return mul(facts[n], mul(modInv(facts[k]), modInv(facts[n-k])));
	}
	public static void main(String[] args) throws IOException 
	{ 
		FastScanner f = new FastScanner(); 
		int t=1;
		t=f.nextInt();
		precomp();
		PrintWriter out=new PrintWriter(System.out);
		for(int tt=0;tt<t;tt++) {
			int n= f.nextInt();
			int[] l=f.readArray(n);
			int and=l[0];
			for(int i=1;i<n;i++) {
				and=and&l[i];
			}
			int count=0;
			for(int i=0;i<n;i++) {
				if(l[i]==and) count++;
			}
			System.out.println(mul(mul(nCk(count,2),2),facts[n-2]));
		}
		out.close();
	} 
	static void sort(long[] p) {
        ArrayList<Integer> q = new ArrayList<>();
        for (long i: p) q.add((int) i);
        Collections.sort(q);
        for (int i = 0; i < p.length; i++) p[i] = q.get(i);
    }
    
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		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());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
		long[] readLongArray(int n) {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
	}
} 	
//Some things to notice
//Check for the overflow
//Binary Search
//Bitmask
//runtime error most of the time is due to array index out of range