#include <bits/stdc++.h>
using namespace std;
 
#define ll long long
#define ld long double
#define pb push_back
#define mp make_pair
#define f(i,m,n) for(int i = m; i < n; i++)
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vi vector<int>
#define vl vectaor<ll>
#define vii vector<pii>
#define vll vector<pll>
#define vvi vector<vi>
#define vvl vector<vl>
#define ff first
#define ss second
#define FAST ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);srand(time(NULL));
#define INF LONG_LONG_MAX
#define MINF LONG_LONG_MIN

const int MOD=1000000007;



 ll  power(ll x,ll  n)
 {
      ll res=1;
     while(n)
     {
         if(n&1)
         {
             res=(res*x);
         }
         n>>=1;
         if(n!=0)
         {
              x=(x*x);
         }
        
     }
     return res;
 }
 
 ll  powerMOD(ll x,ll  n)
 {
      ll res=1;
     while(n)
     {
         if(n&1)
         {
             res=(res*x)%MOD;
         }
         n>>=1;
         x=(x*x)%MOD;
     }
     return res;
 }
 

bool mycompare(pair<ll,ll> &a,pair<ll,ll> &b)
{
    if(a.ff==b.ff)
    {
        return a.ss<b.ss;
    }
    return a.ff>b.ff;
}


bool mycompare1(pair<int,int> &a,pair<int,int> &b)
{
    return a.ss<b.ss;
}


ll fact[200000+2]={0};

void solve()
{
    ll i,j,k,n,m,l,x,r;
    cin>>n;
    ll a[n];
    f(i,0,n) cin>>a[i];
    x=INT_MAX;
    f(i,0,n) x=min(x,a[i]);
    k=1;
    l=0;
    f(i,0,n) 
    {
        if(a[i]==x) l++;
        if((x&a[i])!=x) k=0;
    }
    if(!k || l==1) cout<<0<<"\n";
    else
    {
        r=(l*(l-1))%MOD;
        r=(r*powerMOD(2,MOD-2))%MOD;
        r=(r*fact[n-2])%MOD;
        r=(r*2)%MOD;
        cout<<r<<"\n";
    }
}

int main()
{
    FAST;
    ll t=1,i;
    fact[0]=1;
    f(i,1,200000+2) fact[i]=(fact[i-1]*i)%MOD;
    cin>>t;
    f(i,0,t) solve();
    return 0;
}
### Prompt

Create a solution in CPP for 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 ll long long
#define ld long double
#define pb push_back
#define mp make_pair
#define f(i,m,n) for(int i = m; i < n; i++)
#define pii pair<int, int>
#define pll pair<ll, ll>
#define vi vector<int>
#define vl vectaor<ll>
#define vii vector<pii>
#define vll vector<pll>
#define vvi vector<vi>
#define vvl vector<vl>
#define ff first
#define ss second
#define FAST ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);srand(time(NULL));
#define INF LONG_LONG_MAX
#define MINF LONG_LONG_MIN

const int MOD=1000000007;



 ll  power(ll x,ll  n)
 {
      ll res=1;
     while(n)
     {
         if(n&1)
         {
             res=(res*x);
         }
         n>>=1;
         if(n!=0)
         {
              x=(x*x);
         }
        
     }
     return res;
 }
 
 ll  powerMOD(ll x,ll  n)
 {
      ll res=1;
     while(n)
     {
         if(n&1)
         {
             res=(res*x)%MOD;
         }
         n>>=1;
         x=(x*x)%MOD;
     }
     return res;
 }
 

bool mycompare(pair<ll,ll> &a,pair<ll,ll> &b)
{
    if(a.ff==b.ff)
    {
        return a.ss<b.ss;
    }
    return a.ff>b.ff;
}


bool mycompare1(pair<int,int> &a,pair<int,int> &b)
{
    return a.ss<b.ss;
}


ll fact[200000+2]={0};

void solve()
{
    ll i,j,k,n,m,l,x,r;
    cin>>n;
    ll a[n];
    f(i,0,n) cin>>a[i];
    x=INT_MAX;
    f(i,0,n) x=min(x,a[i]);
    k=1;
    l=0;
    f(i,0,n) 
    {
        if(a[i]==x) l++;
        if((x&a[i])!=x) k=0;
    }
    if(!k || l==1) cout<<0<<"\n";
    else
    {
        r=(l*(l-1))%MOD;
        r=(r*powerMOD(2,MOD-2))%MOD;
        r=(r*fact[n-2])%MOD;
        r=(r*2)%MOD;
        cout<<r<<"\n";
    }
}

int main()
{
    FAST;
    ll t=1,i;
    fact[0]=1;
    f(i,1,200000+2) fact[i]=(fact[i-1]*i)%MOD;
    cin>>t;
    f(i,0,t) solve();
    return 0;
}
```