#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 2e5 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;

int t, n, k;
int a[MAXN];
LL fact[MAXN];

LL solve(int min_x, int min_num) {
    sort(a+1, a+1+n);
    int x = a[n];
    for (int i = n; i >= 1; i--) {
        if (x < min_x) return 0;
        x &= a[i];
    }
    return 1ll*min_num*(min_num-1) % mod*fact[n-2] % mod;
}

void init() {
    fact[0] = 1;
    for (int i = 1; i <= 2e5; i++) {
        fact[i] = fact[i-1]*i % mod;
    }
}

int main() {
    init();
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        int min_x = inf, min_num = 0;
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            if (a[i] < min_x) min_x = a[i], min_num = 1;
            else if (a[i] == min_x) min_num++;
        }
        printf("%lld\n", solve(min_x, min_num));
    }
}

### 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;
const int MAXN = 2e5 + 10;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;

int t, n, k;
int a[MAXN];
LL fact[MAXN];

LL solve(int min_x, int min_num) {
    sort(a+1, a+1+n);
    int x = a[n];
    for (int i = n; i >= 1; i--) {
        if (x < min_x) return 0;
        x &= a[i];
    }
    return 1ll*min_num*(min_num-1) % mod*fact[n-2] % mod;
}

void init() {
    fact[0] = 1;
    for (int i = 1; i <= 2e5; i++) {
        fact[i] = fact[i-1]*i % mod;
    }
}

int main() {
    init();
    scanf("%d", &t);
    while(t--) {
        scanf("%d", &n);
        int min_x = inf, min_num = 0;
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            if (a[i] < min_x) min_x = a[i], min_num = 1;
            else if (a[i] == min_x) min_num++;
        }
        printf("%lld\n", solve(min_x, min_num));
    }
}

```