#include<bits/stdc++.h>
#define llint long long int
#define all(v) v.begin(),v.end()
#define Fill(a,b)  memset(a,b,sizeof(a))
#define what_is(x) cerr<<#x<<"is"<<x<<'\n';
#define google cout<<"Case #"<<ll<<": "<<day<<'\n';
#define boost std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
using namespace std;
#define mod int(1e9+7)
llint fact[200005];
void fc(){
    fact[0]=1;
    fact[1]=1;
    llint ans=1;
    for(int i=2;i<200003;i++)
    {
        ans*=i;
        ans%=mod;
        fact[i]=ans;
    }
}
llint cf(llint count)
{
    return ((count%mod)*((count-1)%mod))/2;
}
int main(){
    boost
    fc();
    int t;
    cin>>t;
    while(t--)
    {
        llint n;
        cin>>n;
        llint a[n];
        llint mi=10000000000;
        bool possible=true;
        for(int i=0;i<n;i++)
            cin>>a[i],mi=min(mi,a[i]);
        llint count=0;
        for(int i=0;i<n;i++)
            if(a[i]==mi)
                count++;

        if(count<2)
            possible=false;
        else 
        {   //cout<<"hf";
            for(int i=0;i<n;i++)
            if((a[i] & mi)!=mi)
            {
                possible=false;
                //cout<<"fhsf";
                break;
            }

        }  
        if(!possible)
            cout<<0<<'\n';
        else 
        {
            llint ans=1;
            ans=((2*cf(count)%mod)*(fact[n-2]%mod))%mod;
            cout<<ans<<'\n';
        }

    }
    return 0;
}

### Prompt

Your challenge is to write a Cpp solution 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 llint long long int
#define all(v) v.begin(),v.end()
#define Fill(a,b)  memset(a,b,sizeof(a))
#define what_is(x) cerr<<#x<<"is"<<x<<'\n';
#define google cout<<"Case #"<<ll<<": "<<day<<'\n';
#define boost std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
using namespace std;
#define mod int(1e9+7)
llint fact[200005];
void fc(){
    fact[0]=1;
    fact[1]=1;
    llint ans=1;
    for(int i=2;i<200003;i++)
    {
        ans*=i;
        ans%=mod;
        fact[i]=ans;
    }
}
llint cf(llint count)
{
    return ((count%mod)*((count-1)%mod))/2;
}
int main(){
    boost
    fc();
    int t;
    cin>>t;
    while(t--)
    {
        llint n;
        cin>>n;
        llint a[n];
        llint mi=10000000000;
        bool possible=true;
        for(int i=0;i<n;i++)
            cin>>a[i],mi=min(mi,a[i]);
        llint count=0;
        for(int i=0;i<n;i++)
            if(a[i]==mi)
                count++;

        if(count<2)
            possible=false;
        else 
        {   //cout<<"hf";
            for(int i=0;i<n;i++)
            if((a[i] & mi)!=mi)
            {
                possible=false;
                //cout<<"fhsf";
                break;
            }

        }  
        if(!possible)
            cout<<0<<'\n';
        else 
        {
            llint ans=1;
            ans=((2*cf(count)%mod)*(fact[n-2]%mod))%mod;
            cout<<ans<<'\n';
        }

    }
    return 0;
}

```