#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
#include "bits/stdc++.h"
using namespace std;
# define ll  long long int
# define ld  long double
# define pb push_back
typedef priority_queue<ll, vector<ll>, greater<ll>> minheap;
typedef priority_queue<ll> maxheap;
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

const ll M = 1e9 + 7, N = 2e5;
ll fact[N + 1];

void init(){
    fact[0] = fact[1] = 1;
    for(ll i = 2; i <= N; i++)
        fact[i] = (fact[i - 1]*i)%M;
}

void solve(){
    ll n;
    cin >> n;
    ll AND;
    ll arr[n];
    map<ll, ll> mp;
    for(ll i = 0; i < n; i++){
        cin >> arr[i];
        mp[arr[i]]++;
        if(i == 0)
            AND = arr[i];
        else
            AND = AND & arr[i];  
    }
    // if(mp.find(AND) == mp.end() || mp[AND] < 2){
    //     cout << 0 << "\n";
    //     return;
    // }
    ll ans = mp[AND]*(mp[AND] - 1)%M*fact[n - 2]%M;
    cout << ans << "\n";
}

int main()
{
    #ifndef ONLINE_JUDGE
      freopen("input.txt", "r", stdin);
      freopen("output.txt", "w", stdout);
    #endif
     IOS
     int t = 1;
     cin >> t;
     init();
     while(t--){
        solve();
        // cout << fixed << setprecision(7); //For trailing 0's
     }
    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
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
#include "bits/stdc++.h"
using namespace std;
# define ll  long long int
# define ld  long double
# define pb push_back
typedef priority_queue<ll, vector<ll>, greater<ll>> minheap;
typedef priority_queue<ll> maxheap;
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

const ll M = 1e9 + 7, N = 2e5;
ll fact[N + 1];

void init(){
    fact[0] = fact[1] = 1;
    for(ll i = 2; i <= N; i++)
        fact[i] = (fact[i - 1]*i)%M;
}

void solve(){
    ll n;
    cin >> n;
    ll AND;
    ll arr[n];
    map<ll, ll> mp;
    for(ll i = 0; i < n; i++){
        cin >> arr[i];
        mp[arr[i]]++;
        if(i == 0)
            AND = arr[i];
        else
            AND = AND & arr[i];  
    }
    // if(mp.find(AND) == mp.end() || mp[AND] < 2){
    //     cout << 0 << "\n";
    //     return;
    // }
    ll ans = mp[AND]*(mp[AND] - 1)%M*fact[n - 2]%M;
    cout << ans << "\n";
}

int main()
{
    #ifndef ONLINE_JUDGE
      freopen("input.txt", "r", stdin);
      freopen("output.txt", "w", stdout);
    #endif
     IOS
     int t = 1;
     cin >> t;
     init();
     while(t--){
        solve();
        // cout << fixed << setprecision(7); //For trailing 0's
     }
    return 0;
}
```