#include<bits/stdc++.h>
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
using namespace std;
// typedef tree<pair<int,int>, null_type, less<pair<int,int>>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
#define ll long long
#define scn(n) scanf("%d",&n)
#define lscn(n) scanf("%lld",&n)
#define lpri(n) printf("%lld",n)
#define pri(n) printf("%d",n)
#define pln() printf("\n")
#define priln(n) printf("%d\n",n)
#define lpriln(n) printf("%lld\n",n)
#define rep(i,init,n) for(int i=init;i<n;i++)
#define pb push_back     
#define mp make_pair
#define F first
#define S second
#define gcd __gcd
#define inf INT_MAX
#define ninf INT_MIN   
#define inf INT_MAX
#define lmax LLONG_MAX
#define lmin LLONG_MIN   
#define tc ll ttt; cin>>ttt; while(ttt--)
#define VC vector<ll> 
#define MP map<ll,ll> 
#define lb lower_bound
#define ub upper_bound
#define sort(v)   sort(v.begin(),v.end())
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)

const int mod = 1e9 + 7;
const int N = 1e7 + 4;
// here just we have to fix twp corner points with value having equal to the &  value of the array, to make array good.
// so our ans will be =(No of choices at the corners)*(n-2)!.

void solve()
{
   ll n,val; cin >>n;
   ll a[n];
   rep(i,0,n)
   {    
      cin >> a[i];
      if(i==0)
      	val=a[i];

      val=val&a[i];
   }
    ll c=0;

    for(ll i=0;i<n;i++)
    {
    	if(a[i]==val)
    		c++;
    }
      if(c<2)     //here no two elements can be kept to make the array good.
      {
      	cout<<"0"<<endl;
      	return;
      }

      ll ans=1;

      rep(i,1,n-1)
      {
      	ans=(ans*i)%mod;
      }
     c=(c*(c-1))%mod;    // if there are c elemets which can be kept the corner 
                        //  than number of ways will be c*(c-1);

     cout<<(ans*c)%mod<<endl;


}

 
int main(){fast;tc{solve();}}


 
### Prompt

Develop a solution in CPP 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 <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
using namespace std;
// typedef tree<pair<int,int>, null_type, less<pair<int,int>>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
#define ll long long
#define scn(n) scanf("%d",&n)
#define lscn(n) scanf("%lld",&n)
#define lpri(n) printf("%lld",n)
#define pri(n) printf("%d",n)
#define pln() printf("\n")
#define priln(n) printf("%d\n",n)
#define lpriln(n) printf("%lld\n",n)
#define rep(i,init,n) for(int i=init;i<n;i++)
#define pb push_back     
#define mp make_pair
#define F first
#define S second
#define gcd __gcd
#define inf INT_MAX
#define ninf INT_MIN   
#define inf INT_MAX
#define lmax LLONG_MAX
#define lmin LLONG_MIN   
#define tc ll ttt; cin>>ttt; while(ttt--)
#define VC vector<ll> 
#define MP map<ll,ll> 
#define lb lower_bound
#define ub upper_bound
#define sort(v)   sort(v.begin(),v.end())
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL)

const int mod = 1e9 + 7;
const int N = 1e7 + 4;
// here just we have to fix twp corner points with value having equal to the &  value of the array, to make array good.
// so our ans will be =(No of choices at the corners)*(n-2)!.

void solve()
{
   ll n,val; cin >>n;
   ll a[n];
   rep(i,0,n)
   {    
      cin >> a[i];
      if(i==0)
      	val=a[i];

      val=val&a[i];
   }
    ll c=0;

    for(ll i=0;i<n;i++)
    {
    	if(a[i]==val)
    		c++;
    }
      if(c<2)     //here no two elements can be kept to make the array good.
      {
      	cout<<"0"<<endl;
      	return;
      }

      ll ans=1;

      rep(i,1,n-1)
      {
      	ans=(ans*i)%mod;
      }
     c=(c*(c-1))%mod;    // if there are c elemets which can be kept the corner 
                        //  than number of ways will be c*(c-1);

     cout<<(ans*c)%mod<<endl;


}

 
int main(){fast;tc{solve();}}


 
```