#include <bits/stdc++.h>
#include <iostream>
#include <limits.h>
#include <math.h>
#include <cmath>
#define s_mod 1000000007
#define pi 3.14159265
//  sin((deg*pi)/180.0);

using namespace std;

long long int ceil2(long long int a, long long int b) 
{
    return (a + b - 1) / b;
}

int hcf(int n1, int n2)
{
    if (n2 != 0)
       return hcf(n2, n1 % n2);
    else 
       return n1;
}

//                                          人生で最も重要なことは、コードを実行する必要があるということです。

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int t;
    cin>>t;
    
    while(t--)
    {
        long long int n;
        int input;
        vector<long long int> ice;
        
        cin>>n;
        for(int i=0;i<n;i++)
        {
            cin>>input;
            ice.push_back(input);
        }
        
        long long int andx = ice[0];
        for(int i=1;i<n;i++)
        {
            andx = andx&(ice[i]);
        }
        
        //  count same as final value 
        
        long long int count = 0;
        for(int i=0;i<n;i++)
        {
            if(ice[i]==andx)
            {
                count++;    //  will be used in the permutation
            }
        }
        
        long long int permute = 0;
        
        if(count>1)    
        {
            permute = (count*(count-1))%s_mod;
            
            for(int i=(n-2);i>=1;i--)
            {
                permute = (permute*i)%s_mod;
            }
        }
        
        cout<<permute<<endl;
        
        ice.clear();
    }
    return 0;
}
### Prompt

Your task is to create a cpp solution 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>
#include <iostream>
#include <limits.h>
#include <math.h>
#include <cmath>
#define s_mod 1000000007
#define pi 3.14159265
//  sin((deg*pi)/180.0);

using namespace std;

long long int ceil2(long long int a, long long int b) 
{
    return (a + b - 1) / b;
}

int hcf(int n1, int n2)
{
    if (n2 != 0)
       return hcf(n2, n1 % n2);
    else 
       return n1;
}

//                                          人生で最も重要なことは、コードを実行する必要があるということです。

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    
    int t;
    cin>>t;
    
    while(t--)
    {
        long long int n;
        int input;
        vector<long long int> ice;
        
        cin>>n;
        for(int i=0;i<n;i++)
        {
            cin>>input;
            ice.push_back(input);
        }
        
        long long int andx = ice[0];
        for(int i=1;i<n;i++)
        {
            andx = andx&(ice[i]);
        }
        
        //  count same as final value 
        
        long long int count = 0;
        for(int i=0;i<n;i++)
        {
            if(ice[i]==andx)
            {
                count++;    //  will be used in the permutation
            }
        }
        
        long long int permute = 0;
        
        if(count>1)    
        {
            permute = (count*(count-1))%s_mod;
            
            for(int i=(n-2);i>=1;i--)
            {
                permute = (permute*i)%s_mod;
            }
        }
        
        cout<<permute<<endl;
        
        ice.clear();
    }
    return 0;
}
```