#include<bits/stdc++.h>
using namespace std;

#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define int long long
#define FIN(i,n) for(int i=0;i<n;i++)
#define loop(i, a, b) for (int i = a; i < b; i++)
#define vp vector<pair < int, int > >
#define vll vector<long long>
#define F first
#define S second
#define MINV(v) *min_element(v.begin(),v.end())
#define MINIV(v) min_element(v.begin(),v.end())-v.begin()
#define all(v) v.begin(), v.end()
#define pb push_back
#define TR(container, it) for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)

// make_pair()
//vector<vector<char>> v(r, vector<char> (c,0));
int mod=1e9+7;
int INF=1e18;

int fact[200006];
int invfact[200006];
const int N=2e5+5;
int naturalNumInverse[N+1];

bool isPrime(int n){if(n<=1)return false;if(n<=3)return true;
    if(n%2==0||n%3==0)return false;for(int i=5;i*i<=n;i=i+6)
    if(n%i==0||n%(i+2)==0)return false;return true;}

int nCr(int n,int r){int res=1;if(n<r)return -1;r=min(r,n-r);
    return ((fact[n]*invfact[r])%mod *invfact[n-r])%mod;
}

int nPr(int n,int r){int res=1;if(n<r)return -1;
    loop(i,0,r)res*=(n-i);return res;}


int modexp(int x,int y){
    int res=1;
    while(y>1){
        if(y%2==1){
            res = (res*x)%mod;
        }
        x=(x*x)%mod;
        y/=2;
    }
    return res;
}
void solve()
{
    int n;
    cin>>n;
    vll v(n);
    FIN(i,n) cin>>v[i];
    int req=v[0]&v[1];
    for(int i=2;i<n;i++){
        req = req & v[i];
    }
    int ctr=0;
    for(int i=0;i<n;i++)
        if(v[i]==req) ctr++;
    if(ctr<2){
        cout<<0<<"\n";
        return;
    }
    int ans=1;
    ans=(ans*nCr(ctr,2))%mod;
    ans=(ans*2)%mod;
    ans=(ans*fact[n-2])%mod;
    cout<<ans<<"\n";


}

signed main(){
    fast
    fact[0]=1;
    fact[1]=1;
    invfact[0]=1;
    invfact[1]=1;
    for(int i=2;i<200006;i++){
        fact[i]=(fact[i-1]*i)%mod;
    }
    naturalNumInverse[0] = naturalNumInverse[1] = 1;
    for (int i = 2; i <= N; i++)
        naturalNumInverse[i] = naturalNumInverse[mod % i] * (mod - mod / i) % mod;
    for(int i=2;i<200006;i++){
        invfact[i]=(invfact[i-1]*naturalNumInverse[i])%mod;
    }
    // cout<<fact[4]<<' ';
    // cout<<invfact[4]<<' ';
    // cout<<nCr(,4)<<' ';
    int t=1;
    cin>>t;
    while(t--)
        solve();
    return 0;
}
### 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>
using namespace std;

#define fast ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define int long long
#define FIN(i,n) for(int i=0;i<n;i++)
#define loop(i, a, b) for (int i = a; i < b; i++)
#define vp vector<pair < int, int > >
#define vll vector<long long>
#define F first
#define S second
#define MINV(v) *min_element(v.begin(),v.end())
#define MINIV(v) min_element(v.begin(),v.end())-v.begin()
#define all(v) v.begin(), v.end()
#define pb push_back
#define TR(container, it) for(typeof(container.begin()) it = container.begin(); it != container.end(); it++)

// make_pair()
//vector<vector<char>> v(r, vector<char> (c,0));
int mod=1e9+7;
int INF=1e18;

int fact[200006];
int invfact[200006];
const int N=2e5+5;
int naturalNumInverse[N+1];

bool isPrime(int n){if(n<=1)return false;if(n<=3)return true;
    if(n%2==0||n%3==0)return false;for(int i=5;i*i<=n;i=i+6)
    if(n%i==0||n%(i+2)==0)return false;return true;}

int nCr(int n,int r){int res=1;if(n<r)return -1;r=min(r,n-r);
    return ((fact[n]*invfact[r])%mod *invfact[n-r])%mod;
}

int nPr(int n,int r){int res=1;if(n<r)return -1;
    loop(i,0,r)res*=(n-i);return res;}


int modexp(int x,int y){
    int res=1;
    while(y>1){
        if(y%2==1){
            res = (res*x)%mod;
        }
        x=(x*x)%mod;
        y/=2;
    }
    return res;
}
void solve()
{
    int n;
    cin>>n;
    vll v(n);
    FIN(i,n) cin>>v[i];
    int req=v[0]&v[1];
    for(int i=2;i<n;i++){
        req = req & v[i];
    }
    int ctr=0;
    for(int i=0;i<n;i++)
        if(v[i]==req) ctr++;
    if(ctr<2){
        cout<<0<<"\n";
        return;
    }
    int ans=1;
    ans=(ans*nCr(ctr,2))%mod;
    ans=(ans*2)%mod;
    ans=(ans*fact[n-2])%mod;
    cout<<ans<<"\n";


}

signed main(){
    fast
    fact[0]=1;
    fact[1]=1;
    invfact[0]=1;
    invfact[1]=1;
    for(int i=2;i<200006;i++){
        fact[i]=(fact[i-1]*i)%mod;
    }
    naturalNumInverse[0] = naturalNumInverse[1] = 1;
    for (int i = 2; i <= N; i++)
        naturalNumInverse[i] = naturalNumInverse[mod % i] * (mod - mod / i) % mod;
    for(int i=2;i<200006;i++){
        invfact[i]=(invfact[i-1]*naturalNumInverse[i])%mod;
    }
    // cout<<fact[4]<<' ';
    // cout<<invfact[4]<<' ';
    // cout<<nCr(,4)<<' ';
    int t=1;
    cin>>t;
    while(t--)
        solve();
    return 0;
}
```