#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define ll long long
#define mod 1000000007
using namespace std;
void factorial(ll int *fact)
{
	fact[0]=1;
    for(ll int i=1;i<=200000;i++)
    {
    	fact[i]=(fact[i-1]*i)%mod;
	}
}
int main() 
{
  	int t=1;
  	cin>>t;
  	ll int fact[200001];
  	factorial(fact);
  	while(t--)
  	{
  		ll int n;
		cin>>n;
		ll int i,j,a[n];
		ll int mi=INT_MAX;
		for(i=0;i<n;i++)
		{
			cin>>a[i];
			mi=min(mi,a[i]);
		}
		ll int c=0;
		ll int an=mi;
		for(i=0;i<n;i++)
		{
			if(a[i]==mi)
			c++;
			an=an&a[i];
		}
		if(an!=mi && c>=2)
		{
			cout<<0<<endl;
		}
		else
		{
			cout<<((((((c*(c-1))/2)%mod)*fact[n-2])%mod)*2)%mod<<endl;
		}
	}
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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>
#define pb push_back
#define mp make_pair
#define ll long long
#define mod 1000000007
using namespace std;
void factorial(ll int *fact)
{
	fact[0]=1;
    for(ll int i=1;i<=200000;i++)
    {
    	fact[i]=(fact[i-1]*i)%mod;
	}
}
int main() 
{
  	int t=1;
  	cin>>t;
  	ll int fact[200001];
  	factorial(fact);
  	while(t--)
  	{
  		ll int n;
		cin>>n;
		ll int i,j,a[n];
		ll int mi=INT_MAX;
		for(i=0;i<n;i++)
		{
			cin>>a[i];
			mi=min(mi,a[i]);
		}
		ll int c=0;
		ll int an=mi;
		for(i=0;i<n;i++)
		{
			if(a[i]==mi)
			c++;
			an=an&a[i];
		}
		if(an!=mi && c>=2)
		{
			cout<<0<<endl;
		}
		else
		{
			cout<<((((((c*(c-1))/2)%mod)*fact[n-2])%mod)*2)%mod<<endl;
		}
	}
}
```