#include <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include<bits/stdc++.h>
using namespace std;

string s,s1,s2;
vector<int>v,v1,v2,v3;
//long long int i,j,k,l,m,n,p,q,z,sum,cnt,mn,mx,t,x,y,mod=1000000007,diff;
//long long int a[100+10],b[100+10];
long long int a[200000+10],pre[200000+10],b[200000+10],suf[200000+10],col[200000+10],last[200000+10],ans[200000+10];
vector<pair<long long int,long long int>> p1,p2;
vector<pair<long long int,pair<long long int,long long int>>> p3;
multiset<long long int> set1[200000+10],set2;
map<long long int, long long int> m1,m2;

//char c;

long long ll_min(long long x,long long y)
{
    if(x>y)
        return y;
    else
        return x;
}

long long ll_max(long long x,long long y)
{
    if(x>y)
        return x;
    else
        return y;
}

long long int brute_force()
{
    return 0;
}

long long int factorial(int n)
{
    long long int M = 1000000007;
 
    long long int f = 1;
    for (int i = 1; i <= n; i++)
        f = (f*i) % M;  // Now f never can
                        // exceed 10^9+7
    return f;
}

int main()
{
    long long int i,j,k,l,m,n,p,q,r,z,sum,cnt,mn,mx,t,x,y,mod=1000000007,diff,c1,c2,c3,c4;
    double f;
    cin>>t;
    //t=1;
    for(z=1;z<=t;z++)
    {
        cin>>n;
        for(i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        p=a[1];
        for(i=2;i<=n;i++)
        {
            p=p&a[i];
        }
        cnt=0;
        for(i=1;i<=n;i++)
        {
            if(a[i]==p)
            {
                cnt++;
            }
        }
        
        q=(cnt)*(cnt-1)/2;
        q=q%mod;
        q=q*factorial(2);
        q=q%mod;
        q=q*factorial(n-2);
        q=q%mod;
        cout<<q<<endl;
    }
    //end of all test cases
    return 0;
}
### Prompt

Create a solution in CPP for 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 <cmath>
#include <cstdio>
#include <vector>
#include <iostream>
#include <algorithm>
#include<bits/stdc++.h>
using namespace std;

string s,s1,s2;
vector<int>v,v1,v2,v3;
//long long int i,j,k,l,m,n,p,q,z,sum,cnt,mn,mx,t,x,y,mod=1000000007,diff;
//long long int a[100+10],b[100+10];
long long int a[200000+10],pre[200000+10],b[200000+10],suf[200000+10],col[200000+10],last[200000+10],ans[200000+10];
vector<pair<long long int,long long int>> p1,p2;
vector<pair<long long int,pair<long long int,long long int>>> p3;
multiset<long long int> set1[200000+10],set2;
map<long long int, long long int> m1,m2;

//char c;

long long ll_min(long long x,long long y)
{
    if(x>y)
        return y;
    else
        return x;
}

long long ll_max(long long x,long long y)
{
    if(x>y)
        return x;
    else
        return y;
}

long long int brute_force()
{
    return 0;
}

long long int factorial(int n)
{
    long long int M = 1000000007;
 
    long long int f = 1;
    for (int i = 1; i <= n; i++)
        f = (f*i) % M;  // Now f never can
                        // exceed 10^9+7
    return f;
}

int main()
{
    long long int i,j,k,l,m,n,p,q,r,z,sum,cnt,mn,mx,t,x,y,mod=1000000007,diff,c1,c2,c3,c4;
    double f;
    cin>>t;
    //t=1;
    for(z=1;z<=t;z++)
    {
        cin>>n;
        for(i=1;i<=n;i++)
        {
            cin>>a[i];
        }
        p=a[1];
        for(i=2;i<=n;i++)
        {
            p=p&a[i];
        }
        cnt=0;
        for(i=1;i<=n;i++)
        {
            if(a[i]==p)
            {
                cnt++;
            }
        }
        
        q=(cnt)*(cnt-1)/2;
        q=q%mod;
        q=q*factorial(2);
        q=q%mod;
        q=q*factorial(n-2);
        q=q%mod;
        cout<<q<<endl;
    }
    //end of all test cases
    return 0;
}
```