#include <bits/stdc++.h>
#define ull signed long long int
#define pb emplace_back
#define mod 1000000007
#define repi(x,y,z) for(ull i=x;i<=y;i+=z)
#define repj(x,y,z) for(ull j=x;j<=y;j+=z)
using namespace std;

ull nCr(ull n, ull r, ull p)
{
    // Optimization for the cases when r is large
    if (r > n - r)
        r = n - r;
 
    ull C[r + 1];
    memset(C, 0, sizeof(C));
 
    C[0] = 1; 
    for (ull i = 1; i <= n; i++) {
 
       
        for (ull j = min(i, r); j > 0; j--)
 
         
            C[j] = (C[j] + C[j - 1]) % p;
    }
    return C[r];
}





int main()
{
   ios_base::sync_with_stdio(false);
   cin.tie(NULL);
   
   ull tc;
   cin>>tc;
   while(tc--)
   {
   
       ull n;
       cin>>n;
       ull a[n];
       bool flag=true;
       ull mini=1e15;
       repi(0,n-1,1)
       {
        cin>>a[i];
        mini=min(mini,a[i]);
       }
       //cout<<count(a,a+n,mini)<<endl;

       ull cm=0;

       repi(0,n-1,1)
       {
        if(a[i]==mini) cm++;
         if((mini & a[i])!=mini)
         {
            
            flag=false; 
         }
       }
       if(cm<2)
       {
        flag=false;
       }
       //cout<<flag<<

       if(!flag)
       {
        cout<<0<<endl;
        continue;
       }

       ull fact[n+1];
       fact[0]=1;
       repi(1,n-1,1)
       {
        fact[i]=((fact[i-1]*i)%mod);
       }

       cout<<(nCr(cm,2,mod)%mod)*2%mod*(fact[n-2]%mod)%mod<<endl;
      

    }   

return 0;	
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 ull signed long long int
#define pb emplace_back
#define mod 1000000007
#define repi(x,y,z) for(ull i=x;i<=y;i+=z)
#define repj(x,y,z) for(ull j=x;j<=y;j+=z)
using namespace std;

ull nCr(ull n, ull r, ull p)
{
    // Optimization for the cases when r is large
    if (r > n - r)
        r = n - r;
 
    ull C[r + 1];
    memset(C, 0, sizeof(C));
 
    C[0] = 1; 
    for (ull i = 1; i <= n; i++) {
 
       
        for (ull j = min(i, r); j > 0; j--)
 
         
            C[j] = (C[j] + C[j - 1]) % p;
    }
    return C[r];
}





int main()
{
   ios_base::sync_with_stdio(false);
   cin.tie(NULL);
   
   ull tc;
   cin>>tc;
   while(tc--)
   {
   
       ull n;
       cin>>n;
       ull a[n];
       bool flag=true;
       ull mini=1e15;
       repi(0,n-1,1)
       {
        cin>>a[i];
        mini=min(mini,a[i]);
       }
       //cout<<count(a,a+n,mini)<<endl;

       ull cm=0;

       repi(0,n-1,1)
       {
        if(a[i]==mini) cm++;
         if((mini & a[i])!=mini)
         {
            
            flag=false; 
         }
       }
       if(cm<2)
       {
        flag=false;
       }
       //cout<<flag<<

       if(!flag)
       {
        cout<<0<<endl;
        continue;
       }

       ull fact[n+1];
       fact[0]=1;
       repi(1,n-1,1)
       {
        fact[i]=((fact[i-1]*i)%mod);
       }

       cout<<(nCr(cm,2,mod)%mod)*2%mod*(fact[n-2]%mod)%mod<<endl;
      

    }   

return 0;	
}

```