#include<bits/stdc++.h>
using namespace std;
#define M 500000
typedef long long ll;
typedef unsigned long long ull;
inline ll read() {
	char ch;
	ll f = 1, res = 0;
	while ((ch = getchar()) < '0' || ch > '9')
		if (ch == '-') f = -1;
	while (ch >= '0' && ch <= '9') {
		res = (res << 1) + (res << 3) + (ch ^ 48);
		ch = getchar();
	}
	return f * res;
}
inline void pt(ll u,ll mod) {
	printf("%lld", u);
	if (mod == 1)
		putchar(' ');
	else if (mod == 2)
		putchar('\n');
}
ll mod=1e9+7;
ll arr[M];
ll p[M];
int main() {
	//freopen("../in", "r", stdin);
	//freopen("../out", "w", stdout);
	p[1]=1;
	p[0]=1;
	for(ll i=2;i<M;i++)
		p[i]=p[i-1]*i%mod;
	ll q = read();
	while (q--) {
		ll n = read();
		ll k = pow(2, 40) + 0.1 - 1;
		for (ll i = 0; i < n; i++) {
			ll t = read();
			arr[i] = t;
			k &= t;
		}
		ll cnt = 0;
		for (ll i = 0; i < n; i++) {
			if (arr[i] == k)
				cnt++;
		}
		ll ans = cnt * (cnt - 1) % mod;
		pt(ans * p[n - 2] % mod, 2);
	}
	return 0;
}
### Prompt

Please create a solution in CPP 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;
#define M 500000
typedef long long ll;
typedef unsigned long long ull;
inline ll read() {
	char ch;
	ll f = 1, res = 0;
	while ((ch = getchar()) < '0' || ch > '9')
		if (ch == '-') f = -1;
	while (ch >= '0' && ch <= '9') {
		res = (res << 1) + (res << 3) + (ch ^ 48);
		ch = getchar();
	}
	return f * res;
}
inline void pt(ll u,ll mod) {
	printf("%lld", u);
	if (mod == 1)
		putchar(' ');
	else if (mod == 2)
		putchar('\n');
}
ll mod=1e9+7;
ll arr[M];
ll p[M];
int main() {
	//freopen("../in", "r", stdin);
	//freopen("../out", "w", stdout);
	p[1]=1;
	p[0]=1;
	for(ll i=2;i<M;i++)
		p[i]=p[i-1]*i%mod;
	ll q = read();
	while (q--) {
		ll n = read();
		ll k = pow(2, 40) + 0.1 - 1;
		for (ll i = 0; i < n; i++) {
			ll t = read();
			arr[i] = t;
			k &= t;
		}
		ll cnt = 0;
		for (ll i = 0; i < n; i++) {
			if (arr[i] == k)
				cnt++;
		}
		ll ans = cnt * (cnt - 1) % mod;
		pt(ans * p[n - 2] % mod, 2);
	}
	return 0;
}
```