#include<bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define IOS ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
 

int32_t main(){
	IOS;
	int t;
	cin>>t;
	while(t--){
		int n;
		cin>>n;
		int a[n];
		int p=1000000007;
		unordered_map<int ,int>m;
		int an=1;
		for(int i=0;i<n;i++){
			cin>>a[i];
			m[a[i]]++;
			
			 // cout<<an<<a[i]<<endl;
		}
		an=a[0]&a[1];
		for(int i=2;i<n;i++){
			an=an&a[i];
		}
		if(m.find(an)==m.end()||m[an]<2){
			cout<<0<<endl;
			continue;
		}
		int f=0;
		for(int i=0;i<n;i++){
			if(an!=an&a[i]){
				f=1;
				break;
			}
		}
		if(f==1){
			cout<<0<<endl;
			continue;
		}
		int re;
		re=(m[an]*(m[an]-1))%p;
		for(int i=n-2;i>=1;i--){
			re=(re*i)%p;
		}
		cout<<re<<endl;
		}
		
		
	
}
### Prompt

Please create a solution in Cpp to the following problem:

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. 

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define pb push_back
#define IOS ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
 

int32_t main(){
	IOS;
	int t;
	cin>>t;
	while(t--){
		int n;
		cin>>n;
		int a[n];
		int p=1000000007;
		unordered_map<int ,int>m;
		int an=1;
		for(int i=0;i<n;i++){
			cin>>a[i];
			m[a[i]]++;
			
			 // cout<<an<<a[i]<<endl;
		}
		an=a[0]&a[1];
		for(int i=2;i<n;i++){
			an=an&a[i];
		}
		if(m.find(an)==m.end()||m[an]<2){
			cout<<0<<endl;
			continue;
		}
		int f=0;
		for(int i=0;i<n;i++){
			if(an!=an&a[i]){
				f=1;
				break;
			}
		}
		if(f==1){
			cout<<0<<endl;
			continue;
		}
		int re;
		re=(m[an]*(m[an]-1))%p;
		for(int i=n-2;i>=1;i--){
			re=(re*i)%p;
		}
		cout<<re<<endl;
		}
		
		
	
}
```