
#include<bits/stdc++.h>
#include<algorithm>
#define int long long
#define ll long long
#define endl "\n"
#define ff first
#define ss second
#define pb push_back
#define ios ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define bs binary_search
#define mp make_pair
#define pf push_front
#define pof pop_front()
#define pob pop_back()
#define lb lower_bound
#define up upper_bound
#define ld long double
#define meme(a,no)  memset(a,no,sizeof(a))
#define all(v)  v.begin(),v.end()
 

using namespace std; 

const int INF=1e18;

const int N=200000+5;

const int mod=1e9+7;

int32_t main(){
  ios;
  int fact[N];
  fact[0]=1;
  fact[1]=1;
  for(int i=2;i<N;i++){
    fact[i]=((fact[i-1]%mod)*(i%mod))%mod;
  }
  int T;
  cin>>T;
  //T=1;
  while(T--){
    int n;
    cin>>n;
    int a[n];
    int i,j,k,l;
    map<int,int>m;
    int ans=0;
    for(i=0;i<n;i++){
      cin>>a[i];
      m[a[i]]++;
      if(i==0){
        ans=a[i];
      }
      else ans=ans&a[i];
      // cout<<i<<" "<<ans<<endl;
    }
    if(m[ans]<=1){
      cout<<0<<endl;
      continue;
    }
    int calc=0;
    calc=fact[n-2];
    calc=((calc%mod)*((m[ans]-1)%mod))%mod;
    calc=((calc%mod)*(m[ans]%mod))%mod;
    cout<<calc<<endl;
  }    
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
#include<algorithm>
#define int long long
#define ll long long
#define endl "\n"
#define ff first
#define ss second
#define pb push_back
#define ios ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define bs binary_search
#define mp make_pair
#define pf push_front
#define pof pop_front()
#define pob pop_back()
#define lb lower_bound
#define up upper_bound
#define ld long double
#define meme(a,no)  memset(a,no,sizeof(a))
#define all(v)  v.begin(),v.end()
 

using namespace std; 

const int INF=1e18;

const int N=200000+5;

const int mod=1e9+7;

int32_t main(){
  ios;
  int fact[N];
  fact[0]=1;
  fact[1]=1;
  for(int i=2;i<N;i++){
    fact[i]=((fact[i-1]%mod)*(i%mod))%mod;
  }
  int T;
  cin>>T;
  //T=1;
  while(T--){
    int n;
    cin>>n;
    int a[n];
    int i,j,k,l;
    map<int,int>m;
    int ans=0;
    for(i=0;i<n;i++){
      cin>>a[i];
      m[a[i]]++;
      if(i==0){
        ans=a[i];
      }
      else ans=ans&a[i];
      // cout<<i<<" "<<ans<<endl;
    }
    if(m[ans]<=1){
      cout<<0<<endl;
      continue;
    }
    int calc=0;
    calc=fact[n-2];
    calc=((calc%mod)*((m[ans]-1)%mod))%mod;
    calc=((calc%mod)*(m[ans]%mod))%mod;
    cout<<calc<<endl;
  }    
}
```