#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define FOR(i,x,y) for(int i=x;i<=y;i++)
#define ROR(i,x,y) for(int i=x;i>=y;i--)
const int maxn=2e5+5;
const int mod=1e9+7;
 
int t,n;
int x;
map<int,ll>mp;
 
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
 
    cin>>t;
    while(t--){
        cin>>n;
        mp.clear();
        int res;
        for(int i=0;i<n;i++){
            cin>>x;
            mp[x]++;
            if(i==0)res=x;
            else res&=x;
        }
 
        if(mp[res]<2){
            cout<<0<<endl;
        }else{
            ll s=1;
            s=mp[res]*(mp[res]-1);
            s=s%mod;
            for(ll i=2;i<=n-2;i++){
                s*=i;
                s%=mod;
            }
            cout<<s<<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;
#define ll long long
#define FOR(i,x,y) for(int i=x;i<=y;i++)
#define ROR(i,x,y) for(int i=x;i>=y;i--)
const int maxn=2e5+5;
const int mod=1e9+7;
 
int t,n;
int x;
map<int,ll>mp;
 
int main(){
    ios::sync_with_stdio(false);
    cin.tie(0),cout.tie(0);
 
    cin>>t;
    while(t--){
        cin>>n;
        mp.clear();
        int res;
        for(int i=0;i<n;i++){
            cin>>x;
            mp[x]++;
            if(i==0)res=x;
            else res&=x;
        }
 
        if(mp[res]<2){
            cout<<0<<endl;
        }else{
            ll s=1;
            s=mp[res]*(mp[res]-1);
            s=s%mod;
            for(ll i=2;i<=n-2;i++){
                s*=i;
                s%=mod;
            }
            cout<<s<<endl;
        }
    }
 
    return 0;
}
```