#pragma GCC optimize("Ofast")
#pragma GCC optimization ("unroll-loops")
 
#include <bits/stdc++.h>
 
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define ins insert
#define sz(x) (int)x.size()
#define dbg(x) cout << x << "\n";
 
const int N = 2e5 + 5;
const long long int mod = 1e9 + 7;
const long long int Mod = 998244353;
const long double Pi = acos(-1);
const long long int Inf = 4e18;
 
using namespace std;
 
void TestCase (){
        long long int n, ans = 0, p = 1;
        cin >> n;
        map <long long int, long long int> q;
        for(long long int i = 1; i<=n - 2; i++) p = (p * i) % mod;
        for(long long int i = 0, x; i<n; i++){
                cin >> x;
                q[x]++;
        }
        if(q.begin()->se > 1){
                long long int k = q.begin()->fi;
                long long int f = (q.begin()->se * (q.begin()->se - 1));
                f = f % mod;
                q.erase(q.begin());
                bool ok = true;
                for(auto &o: q){
                        if((o.fi & k) != k) ok = false;
                }
                if(ok) ans = (p * f) % mod;
        }
        cout << ans;
}
 
int main(){
        IOS;
        int T = 1;
        cin >> T;
        while(T--){
                TestCase();
                cout << "\n";
        }
        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
#pragma GCC optimize("Ofast")
#pragma GCC optimization ("unroll-loops")
 
#include <bits/stdc++.h>
 
#define IOS ios_base::sync_with_stdio(0); cin.tie(0);
#define pb push_back
#define pf push_front
#define fi first
#define se second
#define ins insert
#define sz(x) (int)x.size()
#define dbg(x) cout << x << "\n";
 
const int N = 2e5 + 5;
const long long int mod = 1e9 + 7;
const long long int Mod = 998244353;
const long double Pi = acos(-1);
const long long int Inf = 4e18;
 
using namespace std;
 
void TestCase (){
        long long int n, ans = 0, p = 1;
        cin >> n;
        map <long long int, long long int> q;
        for(long long int i = 1; i<=n - 2; i++) p = (p * i) % mod;
        for(long long int i = 0, x; i<n; i++){
                cin >> x;
                q[x]++;
        }
        if(q.begin()->se > 1){
                long long int k = q.begin()->fi;
                long long int f = (q.begin()->se * (q.begin()->se - 1));
                f = f % mod;
                q.erase(q.begin());
                bool ok = true;
                for(auto &o: q){
                        if((o.fi & k) != k) ok = false;
                }
                if(ok) ans = (p * f) % mod;
        }
        cout << ans;
}
 
int main(){
        IOS;
        int T = 1;
        cin >> T;
        while(T--){
                TestCase();
                cout << "\n";
        }
        return 0;
}
```