/*
#include <algorithm>
#include <iostream>
#include <vector>
#include <string>
#include <queue>
#include <deque>
#include <tuple>

#include <unordered_set>
#include <unordered_map>
#include <set> // lower_bound : <= ; upper_bound : > ; *s.upper_bound(x) de type X
#include <map> // lower_bound : <= ; upper_bound : > ; *m.upper_bound(x) de type pair<X, Y>
#include <stdio.h>
*/

#include <bits/stdc++.h>
#include <numeric>

using namespace std;


#define hmap unordered_map 
#define hset unordered_set 

#define print_var(x) cout << #x << " is " << x << endl;

template<typename T, typename Compare> 
using heap = priority_queue<T, vector<T>, Compare>;

template<typename T> 
T expo(T n, T x){if (x==0) return 1; T k=expo(n, x>>1); return k*k*(x&1 ? n : 1);}

template<typename T> 
T parmi(T k, T n){if (k<0) return 0; if (k==0||k==0)return 1; return (n*parmi(k-1, n-1))/k;}

int64_t MAX = 1e9+7;
int64_t facto(int64_t n, int64_t v){if (n==0)return v; return n*facto(n-1, v) % MAX;}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);

	int t; cin >> t; 

	for (int steps=0; steps<t; steps++){
		int64_t n; cin >> n; 
		vector<int64_t> a(n);
		for (int64_t i(0); i<n; i++){
			cin >> a[i];
		}

		vector<int64_t> b(a);
		vector<int64_t> c(a);
		for (int64_t i(1); i<n; i++)
			b[i] &= b[i-1];
		for (int64_t i(n-2); i>=0; i--)
			c[i] &= c[i+1];


		vector<int64_t> find(0);
		if (c[1] == a[0])
			find.push_back(0);
		if (b[n-2] == a[n-1])
			find.push_back(n-1);
		for (int64_t i=1; i<n-1; i++){
			if (a[i] == (b[i-1] & c[i+1])){
				find.push_back(i);
			}
		}
		if (find.size() < 2)
			cout << 0 << endl; 
		else {
			cout << facto(n-2, int64_t(int64_t(int64_t(find.size()%MAX) * int64_t((find.size()-1)%MAX) % MAX) % MAX)) % MAX << endl; 
		}


	}
	

}



### 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 <algorithm>
#include <iostream>
#include <vector>
#include <string>
#include <queue>
#include <deque>
#include <tuple>

#include <unordered_set>
#include <unordered_map>
#include <set> // lower_bound : <= ; upper_bound : > ; *s.upper_bound(x) de type X
#include <map> // lower_bound : <= ; upper_bound : > ; *m.upper_bound(x) de type pair<X, Y>
#include <stdio.h>
*/

#include <bits/stdc++.h>
#include <numeric>

using namespace std;


#define hmap unordered_map 
#define hset unordered_set 

#define print_var(x) cout << #x << " is " << x << endl;

template<typename T, typename Compare> 
using heap = priority_queue<T, vector<T>, Compare>;

template<typename T> 
T expo(T n, T x){if (x==0) return 1; T k=expo(n, x>>1); return k*k*(x&1 ? n : 1);}

template<typename T> 
T parmi(T k, T n){if (k<0) return 0; if (k==0||k==0)return 1; return (n*parmi(k-1, n-1))/k;}

int64_t MAX = 1e9+7;
int64_t facto(int64_t n, int64_t v){if (n==0)return v; return n*facto(n-1, v) % MAX;}

int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);

	int t; cin >> t; 

	for (int steps=0; steps<t; steps++){
		int64_t n; cin >> n; 
		vector<int64_t> a(n);
		for (int64_t i(0); i<n; i++){
			cin >> a[i];
		}

		vector<int64_t> b(a);
		vector<int64_t> c(a);
		for (int64_t i(1); i<n; i++)
			b[i] &= b[i-1];
		for (int64_t i(n-2); i>=0; i--)
			c[i] &= c[i+1];


		vector<int64_t> find(0);
		if (c[1] == a[0])
			find.push_back(0);
		if (b[n-2] == a[n-1])
			find.push_back(n-1);
		for (int64_t i=1; i<n-1; i++){
			if (a[i] == (b[i-1] & c[i+1])){
				find.push_back(i);
			}
		}
		if (find.size() < 2)
			cout << 0 << endl; 
		else {
			cout << facto(n-2, int64_t(int64_t(int64_t(find.size()%MAX) * int64_t((find.size()-1)%MAX) % MAX) % MAX)) % MAX << endl; 
		}


	}
	

}



```