#include<bits/stdc++.h>
#define int long long
#define ll long long
using namespace std;

const int N = 2e5+5, MOD = 1e9+7;
int fact[N];

ll exp(ll a,ll n){
    if(n==0)    return 1;
    if(n==1)    return a;
    ll ans=exp(a,n/2);
    ans*=ans;ans%=MOD;
    if(n&1) ans*=a,ans%=MOD;
    return ans;
}

int npr(int n,int r){
	return (fact[n] * exp(fact[n-r], MOD-2)) % MOD;
}

void solve(){
	fact[0] = 1;
	for(int i=1;i<N;i++)
		fact[i] = (fact[i-1] * i) % MOD;
	
	int t;cin>>t;
	while(t--){
		int n;cin>>n;
		int a[n];
		for(int i=0;i<n;i++)	cin>>a[i];
		sort(a, a+n);
		if(a[0] == a[n-1]){
			cout<<fact[n]<<"\n";
			continue;
		}

		int set[35]{};
		int fail = 0;
		for(int b=0;b<32;b++){
			int count[2]{};
			for(int i=0;i<n;i++){
				int flag = (a[i]&(1ll<<b) ? 1:0);
				count[flag]++;
			}
			// cout<<b<<" "<<count[0]<<" "<<count[1]<<"\n";
			if(count[0] && count[1]){
				if(count[0] < 2)	fail = 1;
				set[b] = 1;
				// cout<<b<<" b";
			}
		}
		// cout<<"\n";

		int border = 0;
		for(int i=0;i<n;i++){
			int fail = 0;
			for(int bi=0;bi<32;bi++){
				int flag = (a[i]&(1ll<<bi) ? 1:0);
				if(set[bi] && flag){
					fail = 1;
					break;
				}	
			}
			if(!fail)	border+=1;
		}

		// cout<<border<<" check\n";

		int ans = -1;
		if(fail || border < 2)	ans = 0;
		else ans = (npr(border, 2) * fact[n-2]) % MOD;
		cout<<ans<<"\n";
	}
}

signed main(){
	ios_base::sync_with_stdio(false);cin.tie(NULL);
    #ifndef ONLINE_JUDGE    
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout); 
    #endif
   	solve();
    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>
#define int long long
#define ll long long
using namespace std;

const int N = 2e5+5, MOD = 1e9+7;
int fact[N];

ll exp(ll a,ll n){
    if(n==0)    return 1;
    if(n==1)    return a;
    ll ans=exp(a,n/2);
    ans*=ans;ans%=MOD;
    if(n&1) ans*=a,ans%=MOD;
    return ans;
}

int npr(int n,int r){
	return (fact[n] * exp(fact[n-r], MOD-2)) % MOD;
}

void solve(){
	fact[0] = 1;
	for(int i=1;i<N;i++)
		fact[i] = (fact[i-1] * i) % MOD;
	
	int t;cin>>t;
	while(t--){
		int n;cin>>n;
		int a[n];
		for(int i=0;i<n;i++)	cin>>a[i];
		sort(a, a+n);
		if(a[0] == a[n-1]){
			cout<<fact[n]<<"\n";
			continue;
		}

		int set[35]{};
		int fail = 0;
		for(int b=0;b<32;b++){
			int count[2]{};
			for(int i=0;i<n;i++){
				int flag = (a[i]&(1ll<<b) ? 1:0);
				count[flag]++;
			}
			// cout<<b<<" "<<count[0]<<" "<<count[1]<<"\n";
			if(count[0] && count[1]){
				if(count[0] < 2)	fail = 1;
				set[b] = 1;
				// cout<<b<<" b";
			}
		}
		// cout<<"\n";

		int border = 0;
		for(int i=0;i<n;i++){
			int fail = 0;
			for(int bi=0;bi<32;bi++){
				int flag = (a[i]&(1ll<<bi) ? 1:0);
				if(set[bi] && flag){
					fail = 1;
					break;
				}	
			}
			if(!fail)	border+=1;
		}

		// cout<<border<<" check\n";

		int ans = -1;
		if(fail || border < 2)	ans = 0;
		else ans = (npr(border, 2) * fact[n-2]) % MOD;
		cout<<ans<<"\n";
	}
}

signed main(){
	ios_base::sync_with_stdio(false);cin.tie(NULL);
    #ifndef ONLINE_JUDGE    
        freopen("input.txt","r",stdin);
        freopen("output.txt","w",stdout); 
    #endif
   	solve();
    return 0;
}

```