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


//TEMPLATE
#define len(a)            (ll) a.size()
#define ms(a, x)          memset(a, x, sizeof a)
#define bitcount(n)       __builtin_popcountll(n)
#define FR                ios_base::sync_with_stdio(false);cin.tie(NULL)

#define cin1(a)           cin >> a
#define cin2(a, b)        cin >> a >> b
#define cin3(a, b, c)     cin >> a >> b >> c
#define cin4(a, b, c, d)  cin >> a >> b >> c >> d
#define cots(a)           cout << a << " "
#define cot1(a)           cout << a << "\n"
#define cot2(a, b)        cout << a << " " << b << "\n"
#define cot3(a, b, c)     cout << a << " " << b << " " << c << "\n"
#define cot4(a, b, c, d)  cout << a << " " << b << " " << c << " " << d << "\n"
#define cotcase(cs)       cout << "Case " << cs << ": "
#define yes               cout << "Yes\n"
#define no                cout << "No\n"
#define line              cout << "\n"

#define reversed(s)       reverse(s.begin(), s.end())
#define asort(s)          sort(s.begin(), s.end())
#define dsort(s)          sort(s.rbegin(), s.rend())
#define all(s)            s.begin(), s.end()
#define uniq(s)           s.resize(distance(s.begin(),unique(s.begin(), s.end())))
#define found(s, x)       (s.find(x) != s.end())

#define for0(i, n)        for (i = 0; i < n; i++)
#define for1(i, n)        for (i = 1; i <= n; i++)
#define fora(i, a, b)     for (i = a; i <= b; i++)
#define forb(i, b, a)     for (i = b; i >= a; i--)

#define ll                long long
#define pb                push_back
#define mp                make_pair
#define ld                long double
#define pii               pair <ll, ll>
#define piii              pair <ll, pii>
#define F                 first
#define S                 second
#define pi                acos(-1.0)

#define bug(args ...) cerr << __LINE__ << ": ", err(new istringstream(string(#args)), args), cerr << '\n'
void err(istringstream *iss) {}
template<typename T, typename... Args>
void err(istringstream *iss, const T a, const Args... args) {
    string _name;
    *iss >> _name;
    if (_name.back()==',') _name.pop_back();
    cerr << _name << " = " << fixed << setprecision(5) << a << "; ", err(iss, args...);
}

const ll INF = LLONG_MAX;
const ll SZ = 4e5+5;
const ll mod = 1e9+7;



string s, s1, s2;
ll n, k, ans = 0;
ll a[SZ];
//ll b[SZ];

ll ft[SZ];
void cal_fact() {
    ll i;
    ft[0] = 1;

    for(i = 1; i < SZ; i++) {
        ft[i] = (ft[i-1]%mod * i%mod) % mod;
    }
}

ll nc2(ll n) {
    ll res = (n * (n-1)) / 2;
    return res;
}

int main() {
//    FR;
    cal_fact();
    ll cs = 0, tc = 1, x, y, z, i, j, g, p, q, sum = 0, c = 0, t = 0;
//    ll a, b, d;
//    string s, s1, s2;

    cin1(tc);

    while(tc--) {
        cin1(n);
        for1(i, n) cin1(a[i]);

        sort(a+1, a+n+1);

        bool ok = 1;

        x = a[1], c = 0;
        for1(i, n) {
            x &= a[i];
            if(x != a[1]) ok = 0;

            if(a[i] == a[1]) c++;
        }

        if(!ok || c < 2) { cot1(0); continue; }

        ans = ((nc2(c) * 2)%mod * ft[n-2]) % mod;
        cot1(ans);
    }

    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>
using namespace std;


//TEMPLATE
#define len(a)            (ll) a.size()
#define ms(a, x)          memset(a, x, sizeof a)
#define bitcount(n)       __builtin_popcountll(n)
#define FR                ios_base::sync_with_stdio(false);cin.tie(NULL)

#define cin1(a)           cin >> a
#define cin2(a, b)        cin >> a >> b
#define cin3(a, b, c)     cin >> a >> b >> c
#define cin4(a, b, c, d)  cin >> a >> b >> c >> d
#define cots(a)           cout << a << " "
#define cot1(a)           cout << a << "\n"
#define cot2(a, b)        cout << a << " " << b << "\n"
#define cot3(a, b, c)     cout << a << " " << b << " " << c << "\n"
#define cot4(a, b, c, d)  cout << a << " " << b << " " << c << " " << d << "\n"
#define cotcase(cs)       cout << "Case " << cs << ": "
#define yes               cout << "Yes\n"
#define no                cout << "No\n"
#define line              cout << "\n"

#define reversed(s)       reverse(s.begin(), s.end())
#define asort(s)          sort(s.begin(), s.end())
#define dsort(s)          sort(s.rbegin(), s.rend())
#define all(s)            s.begin(), s.end()
#define uniq(s)           s.resize(distance(s.begin(),unique(s.begin(), s.end())))
#define found(s, x)       (s.find(x) != s.end())

#define for0(i, n)        for (i = 0; i < n; i++)
#define for1(i, n)        for (i = 1; i <= n; i++)
#define fora(i, a, b)     for (i = a; i <= b; i++)
#define forb(i, b, a)     for (i = b; i >= a; i--)

#define ll                long long
#define pb                push_back
#define mp                make_pair
#define ld                long double
#define pii               pair <ll, ll>
#define piii              pair <ll, pii>
#define F                 first
#define S                 second
#define pi                acos(-1.0)

#define bug(args ...) cerr << __LINE__ << ": ", err(new istringstream(string(#args)), args), cerr << '\n'
void err(istringstream *iss) {}
template<typename T, typename... Args>
void err(istringstream *iss, const T a, const Args... args) {
    string _name;
    *iss >> _name;
    if (_name.back()==',') _name.pop_back();
    cerr << _name << " = " << fixed << setprecision(5) << a << "; ", err(iss, args...);
}

const ll INF = LLONG_MAX;
const ll SZ = 4e5+5;
const ll mod = 1e9+7;



string s, s1, s2;
ll n, k, ans = 0;
ll a[SZ];
//ll b[SZ];

ll ft[SZ];
void cal_fact() {
    ll i;
    ft[0] = 1;

    for(i = 1; i < SZ; i++) {
        ft[i] = (ft[i-1]%mod * i%mod) % mod;
    }
}

ll nc2(ll n) {
    ll res = (n * (n-1)) / 2;
    return res;
}

int main() {
//    FR;
    cal_fact();
    ll cs = 0, tc = 1, x, y, z, i, j, g, p, q, sum = 0, c = 0, t = 0;
//    ll a, b, d;
//    string s, s1, s2;

    cin1(tc);

    while(tc--) {
        cin1(n);
        for1(i, n) cin1(a[i]);

        sort(a+1, a+n+1);

        bool ok = 1;

        x = a[1], c = 0;
        for1(i, n) {
            x &= a[i];
            if(x != a[1]) ok = 0;

            if(a[i] == a[1]) c++;
        }

        if(!ok || c < 2) { cot1(0); continue; }

        ans = ((nc2(c) * 2)%mod * ft[n-2]) % mod;
        cot1(ans);
    }

    return 0;
}


```