#include<bits/stdc++.h>
using namespace std;
#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)
typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> p2;
#define sz(a) ll(a.size())
ll p[200005],inv[200005],hell=pow(10,9)+7;
ll add(ll x,ll y){
    x+=y;
    if(x>=hell)
        x-=hell;
    return x;
}
ll sub(ll x,ll y){
    x-=y;
    if(x<0)
        x+=hell;
    return x;
}
ll mul(ll x,ll y){
    x=(x*y)%hell;
    return x;
}
ll exp(ll x,ll y){
    ll res=1;
    x%=hell;
    while(y){
        if(y&1)
            res=mul(res,x);
        y>>=1;
        x=mul(x,x);
    }
    return res;
}
ll bin(ll n,ll r){
    if(n<r)
        return 0;
    return mul(p[n],mul(inv[r],inv[n-r]));
}
void pre(){
    p[0]=inv[0]=1;
    for(ll i=1;i<200005;i++){
        p[i]=mul(p[i-1],i);
        inv[i]=exp(p[i],hell-2);
    }
}
void test2311(){
    ll n;
    cin>>n;
    ll lo=hell;
    map<ll,ll> m;
    for(ll i=0;i<n;i++){
        ll x;
        cin>>x;
        m[x]++;
        lo=min(lo,x);
    }
    for(auto it=m.begin();it!=m.end();it++){
        if(((it->first)&lo)!=lo){
            cout<<"0\n";
            return;
        }
    }
    cout<<mul(mul(bin(m[lo],2),2),p[n-2])<<"\n";
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r" , stdin);
    freopen("output.txt", "w", stdout);
    #endif
    clock_t z=clock();
    ll qc=1;
    cin>>qc;
    pre();
    for(ll i=1;i<=qc;i++){
        test2311();
    }
    debug("Total Time:%.4Lf\n",(ld)(clock()-z)/CLOCKS_PER_SEC);
}
### Prompt

In CPP, your task is to solve 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>
using namespace std;
#define debug(...) fprintf(stderr, __VA_ARGS__), fflush(stderr)
typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> p2;
#define sz(a) ll(a.size())
ll p[200005],inv[200005],hell=pow(10,9)+7;
ll add(ll x,ll y){
    x+=y;
    if(x>=hell)
        x-=hell;
    return x;
}
ll sub(ll x,ll y){
    x-=y;
    if(x<0)
        x+=hell;
    return x;
}
ll mul(ll x,ll y){
    x=(x*y)%hell;
    return x;
}
ll exp(ll x,ll y){
    ll res=1;
    x%=hell;
    while(y){
        if(y&1)
            res=mul(res,x);
        y>>=1;
        x=mul(x,x);
    }
    return res;
}
ll bin(ll n,ll r){
    if(n<r)
        return 0;
    return mul(p[n],mul(inv[r],inv[n-r]));
}
void pre(){
    p[0]=inv[0]=1;
    for(ll i=1;i<200005;i++){
        p[i]=mul(p[i-1],i);
        inv[i]=exp(p[i],hell-2);
    }
}
void test2311(){
    ll n;
    cin>>n;
    ll lo=hell;
    map<ll,ll> m;
    for(ll i=0;i<n;i++){
        ll x;
        cin>>x;
        m[x]++;
        lo=min(lo,x);
    }
    for(auto it=m.begin();it!=m.end();it++){
        if(((it->first)&lo)!=lo){
            cout<<"0\n";
            return;
        }
    }
    cout<<mul(mul(bin(m[lo],2),2),p[n-2])<<"\n";
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r" , stdin);
    freopen("output.txt", "w", stdout);
    #endif
    clock_t z=clock();
    ll qc=1;
    cin>>qc;
    pre();
    for(ll i=1;i<=qc;i++){
        test2311();
    }
    debug("Total Time:%.4Lf\n",(ld)(clock()-z)/CLOCKS_PER_SEC);
}
```