
///Bismillahir Rahmanir Rahim
#include<bits/stdc++.h>

#define ll long long
#define pll  pair<ll,ll>

#define ff    first
#define ss    second

#define  mp  make_pair
#define  pb  push_back

#define cy  cout<<"YES"<<endl
#define cn   cout<<"NO"<<endl


using namespace std;

const ll inf=1e18;
const int mod=1e9+7;
const int mod2=2147483647; ///2^31-1
const int M=500005;
const int N=500005;


ll a[M];

inline void normal(ll &a) { a %= mod; if(a<0) a+=mod; }
inline ll modMul(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a * b) % mod; }
inline ll modAdd(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a + b) % mod; }
ll f[N];
void fact()
{
    f[0] = 1;
    for(ll i = 1; i < N; i++) {
        f[i] = modMul(f[i-1], i);
    }
}

int main()
{

    fact();

    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);


    ll t,i,j;
    cin>>t;
    while(t--)
    {
        ll n;cin>>n;

        ll mn=1e9+10;

    for(i=0;i<n;i++)
    {
        cin>>a[i];
    mn=min(a[i],mn);

    }
       int cnt=0;

        for(i=0;i<n;i++) if(a[i]==mn) cnt++;

    bool flag=true;

    for(int i=0;i<n;i++)
    {

        if(a[i]!=mn and (mn&a[i])!=mn)
        {
            flag=false;break;
        }
    }

    if(flag)
    {


    ll ans=modMul(cnt,cnt-1);


    ans =modMul(ans,f[n-2]);

    cout<<ans<<endl;

    }
    else  cout<<0<<endl;




    }

    return 0;

}




### Prompt

Generate a Cpp solution 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

///Bismillahir Rahmanir Rahim
#include<bits/stdc++.h>

#define ll long long
#define pll  pair<ll,ll>

#define ff    first
#define ss    second

#define  mp  make_pair
#define  pb  push_back

#define cy  cout<<"YES"<<endl
#define cn   cout<<"NO"<<endl


using namespace std;

const ll inf=1e18;
const int mod=1e9+7;
const int mod2=2147483647; ///2^31-1
const int M=500005;
const int N=500005;


ll a[M];

inline void normal(ll &a) { a %= mod; if(a<0) a+=mod; }
inline ll modMul(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a * b) % mod; }
inline ll modAdd(ll a, ll b) { a %= mod, b %= mod; normal(a), normal(b); return (a + b) % mod; }
ll f[N];
void fact()
{
    f[0] = 1;
    for(ll i = 1; i < N; i++) {
        f[i] = modMul(f[i-1], i);
    }
}

int main()
{

    fact();

    ios::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);


    ll t,i,j;
    cin>>t;
    while(t--)
    {
        ll n;cin>>n;

        ll mn=1e9+10;

    for(i=0;i<n;i++)
    {
        cin>>a[i];
    mn=min(a[i],mn);

    }
       int cnt=0;

        for(i=0;i<n;i++) if(a[i]==mn) cnt++;

    bool flag=true;

    for(int i=0;i<n;i++)
    {

        if(a[i]!=mn and (mn&a[i])!=mn)
        {
            flag=false;break;
        }
    }

    if(flag)
    {


    ll ans=modMul(cnt,cnt-1);


    ans =modMul(ans,f[n-2]);

    cout<<ans<<endl;

    }
    else  cout<<0<<endl;




    }

    return 0;

}




```