#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
#define rep(i,a,b) for(int i = (a);i <= (b);++i)
#define re_(i,a,b) for(int i = (a);i < (b);++i)
#define dwn(i,a,b) for(int i = (a);i >= (b);--i)

const int SZ = 2e5 + 5;
const int mod = 1e9 + 7;

int n,a[SZ];LL fac[SZ];

LL sum(vector<int> &v){return accumulate(v.begin(),v.end(),0LL);}
void dbg(){puts("");}
template<typename T, typename... R>void dbg(const T &f, const R &... r) {
    cout << f << " ";
    dbg(r...);
}
template<typename Type>inline void read(Type &xx){
    Type f = 1;char ch;xx = 0;
    for(ch = getchar();ch < '0' || ch > '9';ch = getchar()) if(ch == '-') f = -1;
    for(;ch >= '0' && ch <= '9';ch = getchar()) xx = xx * 10 + ch - '0';
    xx *= f;
}

bool jdg(int x){
    rep(i,1,n) if((x & a[i]) != x) return false;
    return true;
}

int main(int argc, char** argv) {
    fac[0] = 1;rep(i,1,SZ-5) fac[i] = fac[i-1] * i % mod;
    int T;read(T);
    while(T--){
        read(n);
        rep(i,1,n) read(a[i]);
        int x = *min_element(a+1,a+n+1);
        if(!jdg(x)) puts("0");
        else{
            int u = 0;
            rep(i,1,n) if(a[i] == x) ++u;
            printf("%lld\n",fac[n-2]*(1LL*u*(u-1)%mod)%mod);
        }
    }
    return 0;
}
### Prompt

Please provide a CPP coded solution to the problem described below:

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;
#define rep(i,a,b) for(int i = (a);i <= (b);++i)
#define re_(i,a,b) for(int i = (a);i < (b);++i)
#define dwn(i,a,b) for(int i = (a);i >= (b);--i)

const int SZ = 2e5 + 5;
const int mod = 1e9 + 7;

int n,a[SZ];LL fac[SZ];

LL sum(vector<int> &v){return accumulate(v.begin(),v.end(),0LL);}
void dbg(){puts("");}
template<typename T, typename... R>void dbg(const T &f, const R &... r) {
    cout << f << " ";
    dbg(r...);
}
template<typename Type>inline void read(Type &xx){
    Type f = 1;char ch;xx = 0;
    for(ch = getchar();ch < '0' || ch > '9';ch = getchar()) if(ch == '-') f = -1;
    for(;ch >= '0' && ch <= '9';ch = getchar()) xx = xx * 10 + ch - '0';
    xx *= f;
}

bool jdg(int x){
    rep(i,1,n) if((x & a[i]) != x) return false;
    return true;
}

int main(int argc, char** argv) {
    fac[0] = 1;rep(i,1,SZ-5) fac[i] = fac[i-1] * i % mod;
    int T;read(T);
    while(T--){
        read(n);
        rep(i,1,n) read(a[i]);
        int x = *min_element(a+1,a+n+1);
        if(!jdg(x)) puts("0");
        else{
            int u = 0;
            rep(i,1,n) if(a[i] == x) ++u;
            printf("%lld\n",fac[n-2]*(1LL*u*(u-1)%mod)%mod);
        }
    }
    return 0;
}
```