#include <bits/stdc++.h>
using namespace std;

#define 	fastio 		ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define		int 		long long
#define		pb 			push_back
#define		vi 			vector<int>
#define		size(x) 	(int)((x).size())
#define		memf(a) 	memset(a,false,sizeof(a))
#define		F 			first
#define 	S 			second
#define 	all(v) 		v.begin(),v.end()
#define 	rep(i,a,b) 	for(int i=a;i<b;i++)
#define		fr(i, n) 	for(int i=0;i<n;i++)
#define		tr(v) 		for(auto it=v.begin();it!=v.end();it = next(it))

const int inf = 1e16; const int N = 1e5 + 1; const int mod = 1e9 + 7;

void solve() {
	int n; cin >> n;
	int fact[n + 1]; fact[0] = fact[1] = 1;
	for (int i = 2; i <= n; i++) {
		fact[i] = (fact[i - 1] * i) % mod;
	}
	int a[n];
	fr(i, n) cin >> a[i];
	int final_and = a[0];
	fr(i, n) {
		final_and = final_and & a[i];
	}

	int cnt = 0;
	fr(i, n) {
		if (a[i] == final_and) cnt++;
	}

	fr(i, n) {
		if ((final_and & a[i]) != final_and) {
			cout << 0 << "\n";
			return;
		}
	}

	if (cnt < 2) {
		cout << "0\n";
		return;
	}

	int ans = (cnt * (cnt - 1)) % mod;
	ans = (fact[n - 2] * ans) % mod;
	cout << ans << "\n";
}

signed main() {
	fastio;
	int t = 1; cin >> t;
	while (t--) solve();
	return 0;
}
### Prompt

Create a solution in Cpp for 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;

#define 	fastio 		ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define		int 		long long
#define		pb 			push_back
#define		vi 			vector<int>
#define		size(x) 	(int)((x).size())
#define		memf(a) 	memset(a,false,sizeof(a))
#define		F 			first
#define 	S 			second
#define 	all(v) 		v.begin(),v.end()
#define 	rep(i,a,b) 	for(int i=a;i<b;i++)
#define		fr(i, n) 	for(int i=0;i<n;i++)
#define		tr(v) 		for(auto it=v.begin();it!=v.end();it = next(it))

const int inf = 1e16; const int N = 1e5 + 1; const int mod = 1e9 + 7;

void solve() {
	int n; cin >> n;
	int fact[n + 1]; fact[0] = fact[1] = 1;
	for (int i = 2; i <= n; i++) {
		fact[i] = (fact[i - 1] * i) % mod;
	}
	int a[n];
	fr(i, n) cin >> a[i];
	int final_and = a[0];
	fr(i, n) {
		final_and = final_and & a[i];
	}

	int cnt = 0;
	fr(i, n) {
		if (a[i] == final_and) cnt++;
	}

	fr(i, n) {
		if ((final_and & a[i]) != final_and) {
			cout << 0 << "\n";
			return;
		}
	}

	if (cnt < 2) {
		cout << "0\n";
		return;
	}

	int ans = (cnt * (cnt - 1)) % mod;
	ans = (fact[n - 2] * ans) % mod;
	cout << ans << "\n";
}

signed main() {
	fastio;
	int t = 1; cin >> t;
	while (t--) solve();
	return 0;
}
```