#include<bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;

#define endl '\n'
#define pb push_back
#define pf push_front
#define fore(i, s, e) for(int i = s; i < e; ++i)
#define forn(i, n) fore(i, 0, n)

const int MOD = 1e9 + 7;



void solve(){
    int n;
    cin >> n;

    vector<int> v(n);
    forn(i, n) cin >> v[i];

    int mn = 2e9;
    forn(i, n){
        mn = min(mn, v[i]);
    }

    int cnt = 0;
    bool ok = true;

    forn(i, n){
        if( v[i] == mn ) cnt++;
        ok &= ((mn & v[i]) == mn);
    }

    if( !ok ) {
        cout << 0 << endl;
        return;
    }

    vector<ll> f(n + 1);
    f[0] = 1LL;

    for(int i = 1; i <= n; ++i)
        f[i] = f[i - 1] * i % MOD;

    ll ans = 1LL;
    ans = 1LL * cnt * (cnt - 1) % MOD;
    ans = 1LL * ans * f[n - 2] % MOD;
    cout << ans << '\n';
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    cin >> T;
    while(T--){
        solve();
    }
    return 0;
}
### Prompt

In CPP, your task is to solve 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;
using ll = long long;
using ld = long double;

#define endl '\n'
#define pb push_back
#define pf push_front
#define fore(i, s, e) for(int i = s; i < e; ++i)
#define forn(i, n) fore(i, 0, n)

const int MOD = 1e9 + 7;



void solve(){
    int n;
    cin >> n;

    vector<int> v(n);
    forn(i, n) cin >> v[i];

    int mn = 2e9;
    forn(i, n){
        mn = min(mn, v[i]);
    }

    int cnt = 0;
    bool ok = true;

    forn(i, n){
        if( v[i] == mn ) cnt++;
        ok &= ((mn & v[i]) == mn);
    }

    if( !ok ) {
        cout << 0 << endl;
        return;
    }

    vector<ll> f(n + 1);
    f[0] = 1LL;

    for(int i = 1; i <= n; ++i)
        f[i] = f[i - 1] * i % MOD;

    ll ans = 1LL;
    ans = 1LL * cnt * (cnt - 1) % MOD;
    ans = 1LL * ans * f[n - 2] % MOD;
    cout << ans << '\n';
}

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    cin >> T;
    while(T--){
        solve();
    }
    return 0;
}
```