#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define me(a,x) memset(a,x,sizeof(a))
#define sf scanf
#define pf printf
#define rep(i,x,y) for(int i=x;i<=y;i++)
#define per(i,x,y) for(int i=x;i>=y;i--)
#define gcd __gcd
#define pb push_back
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define debug(x) cout<<"x="<<x<<endl
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
const ll llINF=9223372036854775807;
const int maxn=2e5+10;
const double pi=acos(-1.0);
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b) {return a/gcd(a,b)*b;}
using namespace std;
ll n,ans;
ll a[maxn];
void solve()
{
    sf("%lld",&n);
    rep(i,1,n){
        sf("%lld",&a[i]);
    }
    ll z=a[1]&a[2];
    rep(i,3,n){
        z=z&a[i];
    }
    ll sum=0;
    rep(i,1,n){
        if(a[i]==z) sum++;
    }
    ans=sum%mod*(sum-1)%mod;
    for(int i=n-2;i>=1;i--){
        ans=ans*i%mod;
    }
    pf("%lld\n",ans);
}

int main()
{
//    IOS;
    int _;
    sf("%d",&_);
    while(_--){
        solve();
    }
    return 0;
}

### Prompt

Your task is to create 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
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define me(a,x) memset(a,x,sizeof(a))
#define sf scanf
#define pf printf
#define rep(i,x,y) for(int i=x;i<=y;i++)
#define per(i,x,y) for(int i=x;i>=y;i--)
#define gcd __gcd
#define pb push_back
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define debug(x) cout<<"x="<<x<<endl
const int mod=1e9+7;
const int INF=0x3f3f3f3f;
const ll llINF=9223372036854775807;
const int maxn=2e5+10;
const double pi=acos(-1.0);
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b) {return a/gcd(a,b)*b;}
using namespace std;
ll n,ans;
ll a[maxn];
void solve()
{
    sf("%lld",&n);
    rep(i,1,n){
        sf("%lld",&a[i]);
    }
    ll z=a[1]&a[2];
    rep(i,3,n){
        z=z&a[i];
    }
    ll sum=0;
    rep(i,1,n){
        if(a[i]==z) sum++;
    }
    ans=sum%mod*(sum-1)%mod;
    for(int i=n-2;i>=1;i--){
        ans=ans*i%mod;
    }
    pf("%lld\n",ans);
}

int main()
{
//    IOS;
    int _;
    sf("%d",&_);
    while(_--){
        solve();
    }
    return 0;
}

```