//------HARSH------//
#include<bits/stdc++.h> 
#include<vector>
#include<string>
#include<algorithm>
using namespace std;

#define int long long 
#define f(i,a,n) for(int i=a;i<n;i++)
#define rf(i,a,n) for(int i=n-1;i>=a;i--)
#define read(arr,n) for(int i=0;i<n;i++)cin>>arr[i];
#define in cin>>
#define out cout<<
#define fr first
#define sc second
#define VC vector<int>
#define VP vector<pair<int,int>>
#define pb push_back
#define mp make_pair
#define PI 3.1415926535897932384626433832

int gcd(int a, int b){
    if (b == 0){return a;}
    else{return gcd(b,a%b);}
}

int lcm(int a, int b){
    return (a*b)/(gcd(a,b));
}

const int MOD=1e9+7;
vector<int> fact(200001,1);

void solve(){
    int n;
    cin>>n;

    int arr[n];
    read(arr,n);
    map<int,int> vec;
    int AND=arr[0];

    vec[arr[0]]++;
    
    f(i,1,n){
        vec[arr[i]]++;
        AND&=arr[i];
    }

    if(vec[AND]<2){cout<<"0\n";return;}

    int temp=(vec[AND]*(vec[AND]-1)/2)%MOD;

    int ans=(fact[n-2]*temp*2)%MOD;

    cout<<ans<<"\n";
}


int32_t main(){
    ios_base::sync_with_stdio(false);cin.tie(NULL);
    cin.tie(0);
    cout.tie(0);
    
    for(int i=1;i<200001;i++){
        fact[i]=(i)*fact[i-1];
        fact[i]%=MOD;
    }

    int t;
    t=1;
    in t;
    while(t--){
        solve();  
    }    
}
### 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
//------HARSH------//
#include<bits/stdc++.h> 
#include<vector>
#include<string>
#include<algorithm>
using namespace std;

#define int long long 
#define f(i,a,n) for(int i=a;i<n;i++)
#define rf(i,a,n) for(int i=n-1;i>=a;i--)
#define read(arr,n) for(int i=0;i<n;i++)cin>>arr[i];
#define in cin>>
#define out cout<<
#define fr first
#define sc second
#define VC vector<int>
#define VP vector<pair<int,int>>
#define pb push_back
#define mp make_pair
#define PI 3.1415926535897932384626433832

int gcd(int a, int b){
    if (b == 0){return a;}
    else{return gcd(b,a%b);}
}

int lcm(int a, int b){
    return (a*b)/(gcd(a,b));
}

const int MOD=1e9+7;
vector<int> fact(200001,1);

void solve(){
    int n;
    cin>>n;

    int arr[n];
    read(arr,n);
    map<int,int> vec;
    int AND=arr[0];

    vec[arr[0]]++;
    
    f(i,1,n){
        vec[arr[i]]++;
        AND&=arr[i];
    }

    if(vec[AND]<2){cout<<"0\n";return;}

    int temp=(vec[AND]*(vec[AND]-1)/2)%MOD;

    int ans=(fact[n-2]*temp*2)%MOD;

    cout<<ans<<"\n";
}


int32_t main(){
    ios_base::sync_with_stdio(false);cin.tie(NULL);
    cin.tie(0);
    cout.tie(0);
    
    for(int i=1;i<200001;i++){
        fact[i]=(i)*fact[i-1];
        fact[i]%=MOD;
    }

    int t;
    t=1;
    in t;
    while(t--){
        solve();  
    }    
}
```