#include <bits/stdc++.h>
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define per(i, a, b) for (int i = a; i >= b; i--)
using namespace std;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef long long ll;

template <typename T>
inline void read(T &f) {
    f = 0; T fu = 1; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') { fu = -1; } c = getchar(); }
    while (c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
    f *= fu;
}

template <typename T>
void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x < 10) putchar(x + 48);
    else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
    print(x); putchar(t);
}
const ll N=2e6+100,mod=1e9+7,inf=0x3f3f3f;
ll t;
ll a[N];
ll A(ll n,ll m) {
	ll res=1;
	for(ll i=m-n+1; i<=m; i++)res=res*i%mod;
	return res;
}
signed main() {
//	cout<<min(inf,1000000000ll)<<endl; 
	read(t);
	while(t--) {
		ll n;
		read(n);
		ll minn=inf;
		map<ll,ll>mp;
		ll jishu=0;ll mx;
		for(ll i=1; i<=n; i++) {
			int a;read(a);
			if(i==1)mx=a;
			else mx=mx&a;
			mp[a]++;
		}
//		cout<<zui<<" "<<mp[zui]<<endl;
		if(mp[mx]>=2){
			ll res=1;
			res*=A(n-2,n-2)%mod;
			res%=mod;
			res*=A(2,mp[mx])%mod;
			res%=mod;
			print(res%mod,'\n');
		}else print(0,'\n');
	}
}

### 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>
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define per(i, a, b) for (int i = a; i >= b; i--)
using namespace std;
typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef long long ll;

template <typename T>
inline void read(T &f) {
    f = 0; T fu = 1; char c = getchar();
    while (c < '0' || c > '9') { if (c == '-') { fu = -1; } c = getchar(); }
    while (c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
    f *= fu;
}

template <typename T>
void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x < 10) putchar(x + 48);
    else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
    print(x); putchar(t);
}
const ll N=2e6+100,mod=1e9+7,inf=0x3f3f3f;
ll t;
ll a[N];
ll A(ll n,ll m) {
	ll res=1;
	for(ll i=m-n+1; i<=m; i++)res=res*i%mod;
	return res;
}
signed main() {
//	cout<<min(inf,1000000000ll)<<endl; 
	read(t);
	while(t--) {
		ll n;
		read(n);
		ll minn=inf;
		map<ll,ll>mp;
		ll jishu=0;ll mx;
		for(ll i=1; i<=n; i++) {
			int a;read(a);
			if(i==1)mx=a;
			else mx=mx&a;
			mp[a]++;
		}
//		cout<<zui<<" "<<mp[zui]<<endl;
		if(mp[mx]>=2){
			ll res=1;
			res*=A(n-2,n-2)%mod;
			res%=mod;
			res*=A(2,mp[mx])%mod;
			res%=mod;
			print(res%mod,'\n');
		}else print(0,'\n');
	}
}

```