#include<bits/stdc++.h>
#define ull unsigned long long int
#define ll long long
#define vi vector<int>
#define vll vector<ll>
#define pb push_back
#define Loop(i,a,b) for (int i=a;i<b;i++)

const int mod=1000000007;

using namespace std;
ll solve2(int n)
{
	ll res=1;
	for (int i=2;i<=n;i++)
	{
		res=(res*i)%mod;
	}
	return res;
}


void solve()
{
	int n;
	cin>>n;
	
	vector<int> vec;
	int tt;
	int mini=INT_MAX;
	for (int i=1;i<=n;i++)
	{
		cin>>tt;
		vec.pb(tt);
		mini=min(mini,tt);
		
	}
	
	int ct=0;
	int flag=0;
	for (int i=0;i<n;i++)
	{	
	
		if (vec[i]&mini!=mini )
		{
			cout<<0<<endl;
			return ;
		}
		if (vec[i]==mini)
		ct++;
	}
	
	if(ct<2 or flag==1)
	{
		cout<<0<<endl;
		return ;
	}
	
	ll val=solve2(n-2);
	ll val2=((ct%mod)*((ct-1)%mod))%mod;
	
	ll ans=((val%mod)*(val2%mod))%mod;
	cout<<ans<<endl;
}

void solveTestCase()
{
    int MOD=1e9+7;
    int n;
    cin>>n;
    vector<int> a(n);
    for(int i=0;i<n;i++)cin>>a[i];
    
    int min1=*min_element(a.begin(),a.end());
    int cnt=0;
    
    for(int x:a)
    {
        if(min1==x)cnt++;
        if((min1&x)!=min1)
        {
            printf("0\n");
            return;
        }
    }
    
    int fact=1;
    for(int i=1;i<=n-2;i++)fact=(1LL*fact*i)%MOD;
    int ans=(1LL * cnt * (cnt-1))%MOD;
    ans = (1LL * ans * fact) % MOD;
    printf("%d\n",ans);
}
int main()
{	
//ios_base::sync_with_stdio(false);
//cin.tie(NULL);
	int t;
	cin>>t;
	while (t--)
	{
	 solveTestCase();
	}
 return 0;
}


### Prompt

In Cpp, your task is to solve 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 ull unsigned long long int
#define ll long long
#define vi vector<int>
#define vll vector<ll>
#define pb push_back
#define Loop(i,a,b) for (int i=a;i<b;i++)

const int mod=1000000007;

using namespace std;
ll solve2(int n)
{
	ll res=1;
	for (int i=2;i<=n;i++)
	{
		res=(res*i)%mod;
	}
	return res;
}


void solve()
{
	int n;
	cin>>n;
	
	vector<int> vec;
	int tt;
	int mini=INT_MAX;
	for (int i=1;i<=n;i++)
	{
		cin>>tt;
		vec.pb(tt);
		mini=min(mini,tt);
		
	}
	
	int ct=0;
	int flag=0;
	for (int i=0;i<n;i++)
	{	
	
		if (vec[i]&mini!=mini )
		{
			cout<<0<<endl;
			return ;
		}
		if (vec[i]==mini)
		ct++;
	}
	
	if(ct<2 or flag==1)
	{
		cout<<0<<endl;
		return ;
	}
	
	ll val=solve2(n-2);
	ll val2=((ct%mod)*((ct-1)%mod))%mod;
	
	ll ans=((val%mod)*(val2%mod))%mod;
	cout<<ans<<endl;
}

void solveTestCase()
{
    int MOD=1e9+7;
    int n;
    cin>>n;
    vector<int> a(n);
    for(int i=0;i<n;i++)cin>>a[i];
    
    int min1=*min_element(a.begin(),a.end());
    int cnt=0;
    
    for(int x:a)
    {
        if(min1==x)cnt++;
        if((min1&x)!=min1)
        {
            printf("0\n");
            return;
        }
    }
    
    int fact=1;
    for(int i=1;i<=n-2;i++)fact=(1LL*fact*i)%MOD;
    int ans=(1LL * cnt * (cnt-1))%MOD;
    ans = (1LL * ans * fact) % MOD;
    printf("%d\n",ans);
}
int main()
{	
//ios_base::sync_with_stdio(false);
//cin.tie(NULL);
	int t;
	cin>>t;
	while (t--)
	{
	 solveTestCase();
	}
 return 0;
}


```