#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<stack>
#include<set>
#define iss ios::sync_with_stdio(false)
using namespace std;
typedef long long ll;
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 7;
const int mod = 1e9 + 7;
const int rr = 233;
int T;
int n;
ll pre[maxn],a[maxn];
ll cal(ll cnt){
    return cnt * (cnt - 1) % mod;
}
int main (){
    scanf("%d",&T);
    pre[0] = 1;
    for(int i = 1; i <= 1e6; i ++) pre[i] = pre[i - 1] * i % mod;
    while(T--){
        scanf("%d",&n);
        int cnt = 0;
        for(int i = 1; i <= n; i ++){
            scanf("%lld",&a[i]);
        }
        ll ans = a[1];
        for(int i = 2; i <= n; i ++){
            ans = ans & a[i];
        }
        for(int i = 1; i <= n; i ++){
            if(a[i] == ans) cnt ++;
        }
        ll pr = cal(cnt) % mod * pre[n - 2] % mod;;
        printf ("%lld\n",pr);



    }






}

### Prompt

Please formulate 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<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<stack>
#include<set>
#define iss ios::sync_with_stdio(false)
using namespace std;
typedef long long ll;
using namespace std;
typedef long long ll;
const int maxn = 1e6 + 7;
const int mod = 1e9 + 7;
const int rr = 233;
int T;
int n;
ll pre[maxn],a[maxn];
ll cal(ll cnt){
    return cnt * (cnt - 1) % mod;
}
int main (){
    scanf("%d",&T);
    pre[0] = 1;
    for(int i = 1; i <= 1e6; i ++) pre[i] = pre[i - 1] * i % mod;
    while(T--){
        scanf("%d",&n);
        int cnt = 0;
        for(int i = 1; i <= n; i ++){
            scanf("%lld",&a[i]);
        }
        ll ans = a[1];
        for(int i = 2; i <= n; i ++){
            ans = ans & a[i];
        }
        for(int i = 1; i <= n; i ++){
            if(a[i] == ans) cnt ++;
        }
        ll pr = cal(cnt) % mod * pre[n - 2] % mod;;
        printf ("%lld\n",pr);



    }






}

```