#include<bits/stdc++.h>
#define ll long long
#define ld long double
#define f first
#define s second
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;
const int mod=1e9+7;
const int maxV=2e5+1;
const ll inf=1e18+1;
int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        vector<ll> a(n);
        for(int i=0;i<n;i++){
            cin>>a[i];
        }
        ll andVal=a[0];
        for(int i=0;i<n;i++){
            andVal=(andVal)&(a[i]);
        }
        ll cnt=0;
        for(int i=0;i<n;i++){
            if(a[i]==andVal){
                cnt++;
            }
        }
        ll ans=(cnt*(cnt-1))%mod;
        for(ll i=1;i<=n-2;i++){
            ans=(ans*i)%mod;
        }
        ans=(ans%mod+mod)%mod;
        cout<<ans<<"\n";
    }
}
### 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 ll long long
#define ld long double
#define f first
#define s second
#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
using namespace std;
const int mod=1e9+7;
const int maxV=2e5+1;
const ll inf=1e18+1;
int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        vector<ll> a(n);
        for(int i=0;i<n;i++){
            cin>>a[i];
        }
        ll andVal=a[0];
        for(int i=0;i<n;i++){
            andVal=(andVal)&(a[i]);
        }
        ll cnt=0;
        for(int i=0;i<n;i++){
            if(a[i]==andVal){
                cnt++;
            }
        }
        ll ans=(cnt*(cnt-1))%mod;
        for(ll i=1;i<=n-2;i++){
            ans=(ans*i)%mod;
        }
        ans=(ans%mod+mod)%mod;
        cout<<ans<<"\n";
    }
}
```