#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> ii;
typedef tuple<int,int,int> i3;

int mod = 1e9+7;
ll fat[200005];

int main(){

    fat[0] = 1;
    for(int i = 1; i <= 2e5; i++)
        fat[i] = fat[i-1] * i % mod;

    cout.sync_with_stdio(0);
    cin.tie(0);
    int t; cin >> t;
    while(t--){
        int n; cin >> n;

        map<int,int> mapa;
        vector<int> v(n);
        for(int i = 0; i < n; i++){
            int x; cin >> x;
            v[i] = x;
            mapa[x]++;
        }
        sort(v.begin(), v.end());
        int aux = 0;
        for(int i = 0; i < n; i++)
            if( (v[0]&v[i]) == v[0])
                aux++;
        int cnt = 0;
        for(auto par : mapa){
            cnt = par.second;
            break;
        }
        if(cnt < 2 || aux != n){
            cout << 0 << endl;
            continue;
        }
        ll res = 1LL * cnt * (cnt - 1) % mod;
        res = res * fat[n-2] % mod;
        cout << res << endl;
    }

	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>
using namespace std;
typedef long long ll;
typedef pair<int,int> ii;
typedef tuple<int,int,int> i3;

int mod = 1e9+7;
ll fat[200005];

int main(){

    fat[0] = 1;
    for(int i = 1; i <= 2e5; i++)
        fat[i] = fat[i-1] * i % mod;

    cout.sync_with_stdio(0);
    cin.tie(0);
    int t; cin >> t;
    while(t--){
        int n; cin >> n;

        map<int,int> mapa;
        vector<int> v(n);
        for(int i = 0; i < n; i++){
            int x; cin >> x;
            v[i] = x;
            mapa[x]++;
        }
        sort(v.begin(), v.end());
        int aux = 0;
        for(int i = 0; i < n; i++)
            if( (v[0]&v[i]) == v[0])
                aux++;
        int cnt = 0;
        for(auto par : mapa){
            cnt = par.second;
            break;
        }
        if(cnt < 2 || aux != n){
            cout << 0 << endl;
            continue;
        }
        ll res = 1LL * cnt * (cnt - 1) % mod;
        res = res * fat[n-2] % mod;
        cout << res << endl;
    }

	return 0;
}
```