#include <bits/stdc++.h>
using namespace std;


int main() {
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        cin>>n;
        long long int a[n];
        int ans;
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
            if(i==0)
            ans=a[i];
            else
            ans=ans&a[i];
        }
        unsigned long long int cnt=0;
        for(int i=0;i<n;i++)
        {
            if(a[i]==ans)
            cnt+=1;
        }
        if(cnt==0||cnt==1)
        {
            cout<<"0"<<endl;
            continue;
        }
        unsigned long long int fact=1;
        for(unsigned long long int i=2;i<=n-2;i++)
        {
            fact=(fact*i)%1000000007;
        }
        unsigned long long int as=((cnt*(cnt-1)%1000000007)*(fact%1000000007));
        cout<<as%1000000007<<endl;
    }
	return 0;
}

### Prompt

Develop a solution in CPP 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>
using namespace std;


int main() {
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int n;
        cin>>n;
        long long int a[n];
        int ans;
        for(int i=0;i<n;i++)
        {
            cin>>a[i];
            if(i==0)
            ans=a[i];
            else
            ans=ans&a[i];
        }
        unsigned long long int cnt=0;
        for(int i=0;i<n;i++)
        {
            if(a[i]==ans)
            cnt+=1;
        }
        if(cnt==0||cnt==1)
        {
            cout<<"0"<<endl;
            continue;
        }
        unsigned long long int fact=1;
        for(unsigned long long int i=2;i<=n-2;i++)
        {
            fact=(fact*i)%1000000007;
        }
        unsigned long long int as=((cnt*(cnt-1)%1000000007)*(fact%1000000007));
        cout<<as%1000000007<<endl;
    }
	return 0;
}

```