#include"bits/stdc++.h"
using namespace std;
#define PB push_back
#define ll long long
const int N=2e5+7;
const ll mod=1e9+7;

ll f[N];

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	int T,cs=0;
	cin>>T;
	f[0]=1;
	for(int i=1;i<N;i++){
		f[i]=(f[i-1]*1ll*i)%mod;
	}
	while(T--){
	    int n;
	    cin>>n;
	    map<int,ll> cnt;
	    int k;
	    for(int i=0;i<n;i++){
	    	int x;
	    	cin>>x;
	    	cnt[x]++;
	    	if(i==0) k=x;
	    	else k&=x;
	    }

	    ll res=(cnt[k]*(cnt[k]-1))%mod;
	    res=(res*f[n-2])%mod;
	    
	    cout<<res<<'\n';
	}
	
		
	return 0;
}
### 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 PB push_back
#define ll long long
const int N=2e5+7;
const ll mod=1e9+7;

ll f[N];

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	int T,cs=0;
	cin>>T;
	f[0]=1;
	for(int i=1;i<N;i++){
		f[i]=(f[i-1]*1ll*i)%mod;
	}
	while(T--){
	    int n;
	    cin>>n;
	    map<int,ll> cnt;
	    int k;
	    for(int i=0;i<n;i++){
	    	int x;
	    	cin>>x;
	    	cnt[x]++;
	    	if(i==0) k=x;
	    	else k&=x;
	    }

	    ll res=(cnt[k]*(cnt[k]-1))%mod;
	    res=(res*f[n-2])%mod;
	    
	    cout<<res<<'\n';
	}
	
		
	return 0;
}
```