#include <bits/stdc++.h>

using namespace std;

// #pragma GCC optimize("Ofast")
// #pragma GCC optimize("fast-math")
// #pragma GCC optimize("unroll-loops")
// #pragma GCC optimize("no-stack-protector")
// #pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx,tune=native")

#define f first
#define s second
#define endl '\n'
#define mp make_pair
#define pb push_back
#define int long long
#define ld long double
#define eb emplace_back
#define sqr(x) (x) * (x)
#define endf endl << flush
#define random() srand(time(NULL))
#define all(x) (x).begin(), (x).end()
#define watch(x) cout << (#x) << " is " << (x) << endf
#define fast() ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL)

const int P = 31;
const int INF = 1e18;
const int base = 1e6;
const int mod = 1e9 + 7;
const double eps = 1e-6;

signed main() {
    random();
    // freopen("input.txt", "r", stdin);
    // freopen("parade.out", "w", stdout);
    fast();
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        int a[n];
        int an = (1 << 30) - 1;
        for (int i = 0; i < n; i++) {
            cin >> a[i];
            an &= a[i];
        }
        int kol = 0;
        for (int i = 0; i < n; i++)
            if (a[i] == an)
                kol++;
        if (kol < 2)
            cout << "0\n";
        else {
            int ans = kol * (kol - 1);
            for (int i = 2; i <= n - 2; i++)
                ans = (ans * i) % mod;
            cout << ans << endl;
        }
    }
    return 0;
}
### Prompt

Generate 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 <bits/stdc++.h>

using namespace std;

// #pragma GCC optimize("Ofast")
// #pragma GCC optimize("fast-math")
// #pragma GCC optimize("unroll-loops")
// #pragma GCC optimize("no-stack-protector")
// #pragma GCC target("sse,sse2,sse3,ssse3,popcnt,abm,mmx,tune=native")

#define f first
#define s second
#define endl '\n'
#define mp make_pair
#define pb push_back
#define int long long
#define ld long double
#define eb emplace_back
#define sqr(x) (x) * (x)
#define endf endl << flush
#define random() srand(time(NULL))
#define all(x) (x).begin(), (x).end()
#define watch(x) cout << (#x) << " is " << (x) << endf
#define fast() ios_base::sync_with_stdio(0); cin.tie(NULL); cout.tie(NULL)

const int P = 31;
const int INF = 1e18;
const int base = 1e6;
const int mod = 1e9 + 7;
const double eps = 1e-6;

signed main() {
    random();
    // freopen("input.txt", "r", stdin);
    // freopen("parade.out", "w", stdout);
    fast();
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        int a[n];
        int an = (1 << 30) - 1;
        for (int i = 0; i < n; i++) {
            cin >> a[i];
            an &= a[i];
        }
        int kol = 0;
        for (int i = 0; i < n; i++)
            if (a[i] == an)
                kol++;
        if (kol < 2)
            cout << "0\n";
        else {
            int ans = kol * (kol - 1);
            for (int i = 2; i <= n - 2; i++)
                ans = (ans * i) % mod;
            cout << ans << endl;
        }
    }
    return 0;
}
```