#include<bits/stdc++.h>
#include<string.h>
#define ll long long
#include<vector>
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
const int mod=1e9+7;
ll fact[200005];
void fat(int n=200005){
	fact[0]=1;
	for(int i=1;i<n;i++){
		fact[i]=(fact[i-1]*i)%mod;
	}
}
int main(){
	IOS;
	int t;
	cin>>t;
	fat();
	while(t--){
		int n;
		cin>>n;
		vector<ll>v(n);
		vector<int> bits(31);
		for(int i=0;i<n;i++){
			cin>>v[i];
			for(int j=0;j<31;j++){
				if(v[i]&(1LL<<j)){
					bits[j]++;
				}
			}	
		}
		ll opt=0;
		for(int i=0;i<n;i++){
			bool can_use=true;
			for(int j=0;j<31;j++){
				if(v[i]&(1LL<<j)){
					if(bits[j]!=n){
						can_use=false;
						break;
					}
				}
			}
			if(can_use){
				opt++;
			}
		}
		ll ans=0;
		ans=(((opt)*(opt-1))/2)%mod;
		ans= (ans*(fact[n-2]))%mod;
		cout<<(ans*2)%mod<<endl;
	}
}
### Prompt

In CPP, your task is to solve 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>
#include<string.h>
#define ll long long
#include<vector>
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
const int mod=1e9+7;
ll fact[200005];
void fat(int n=200005){
	fact[0]=1;
	for(int i=1;i<n;i++){
		fact[i]=(fact[i-1]*i)%mod;
	}
}
int main(){
	IOS;
	int t;
	cin>>t;
	fat();
	while(t--){
		int n;
		cin>>n;
		vector<ll>v(n);
		vector<int> bits(31);
		for(int i=0;i<n;i++){
			cin>>v[i];
			for(int j=0;j<31;j++){
				if(v[i]&(1LL<<j)){
					bits[j]++;
				}
			}	
		}
		ll opt=0;
		for(int i=0;i<n;i++){
			bool can_use=true;
			for(int j=0;j<31;j++){
				if(v[i]&(1LL<<j)){
					if(bits[j]!=n){
						can_use=false;
						break;
					}
				}
			}
			if(can_use){
				opt++;
			}
		}
		ll ans=0;
		ans=(((opt)*(opt-1))/2)%mod;
		ans= (ans*(fact[n-2]))%mod;
		cout<<(ans*2)%mod<<endl;
	}
}
```