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

#define start cin>>t;while(t--)
#define f1 for(ll i=0;i<n;i++)
#define fv for(auto i: v)
#define w1 while(n--)
#define ll long long int
#define pnew printf("\n");
#define prill printf("%lld",
#define vall v.begin(),v.end()
#define so_gr sort(vall,greater<>());


/*vector<int> primes;
bitset<1000000000> mark;
void sieve(int n) {
for(int i = 2; i*i<=n; i++){
    if(mark[i] == false){
        for(int j = i+i; j<=n; j+=i) {
            mark[j] = true;
        }
    }
}
for(int i = 2; i<=n; i++){
    if(mark[i] == false)
        primes.push_back(i);
}
}*/
int main(){
ll t,n,k,in,d,x;
start{
cin>>n;
vector<ll>v;
f1{
cin>>in;
v.push_back(in);
if(i>0){
d=(d&in);
}
else
d=in;
}
ll d1=count(vall,d);
if(d1>1){
x=d1*(d1-1);
x%=1000000007;
ll x1=1;
for(ll i=2;i<=(n-2);i++){
x1*=i;
x1%=1000000007;
}
cout<<((x*x1)%1000000007)<<endl;
}
else{
cout<<0<<endl;
}
}
}

### Prompt

Please create a solution in cpp to 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 start cin>>t;while(t--)
#define f1 for(ll i=0;i<n;i++)
#define fv for(auto i: v)
#define w1 while(n--)
#define ll long long int
#define pnew printf("\n");
#define prill printf("%lld",
#define vall v.begin(),v.end()
#define so_gr sort(vall,greater<>());


/*vector<int> primes;
bitset<1000000000> mark;
void sieve(int n) {
for(int i = 2; i*i<=n; i++){
    if(mark[i] == false){
        for(int j = i+i; j<=n; j+=i) {
            mark[j] = true;
        }
    }
}
for(int i = 2; i<=n; i++){
    if(mark[i] == false)
        primes.push_back(i);
}
}*/
int main(){
ll t,n,k,in,d,x;
start{
cin>>n;
vector<ll>v;
f1{
cin>>in;
v.push_back(in);
if(i>0){
d=(d&in);
}
else
d=in;
}
ll d1=count(vall,d);
if(d1>1){
x=d1*(d1-1);
x%=1000000007;
ll x1=1;
for(ll i=2;i<=(n-2);i++){
x1*=i;
x1%=1000000007;
}
cout<<((x*x1)%1000000007)<<endl;
}
else{
cout<<0<<endl;
}
}
}

```