B. AND Sequences

Divide by Zero 2021 and Codeforces Round #714 (Div. 2)B. AND Sequences

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A sequence of nn non-negative integers (n≥2n≥2) a1,a2,…,ana1,a2,…,an is called good if for all ii from 11 to n−1n−1 the following condition holds true:

a1&a2&…&ai=ai+1&ai+2&…&an,a1&a2&…&ai=ai+1&ai+2&…&an,

where && denotes the bitwise AND operation.

You are given an array aa of size nn (n≥2n≥2). Find the number of permutations pp of numbers ranging from 11 to nn, for which the sequence ap1ap1, ap2ap2, ... ,apnapn is good. Since this number can be large, output it modulo 109+7109+7.

Input

The first line contains a single integer tt (1≤t≤1041≤t≤104), denoting the number of test cases.

The first line of each test case contains a single integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the size of the array.

The second line of each test case contains nn integers a1,a2,…,ana1,a2,…,an (0≤ai≤1090≤ai≤109) — the elements of the array.

It is guaranteed that the sum of nn over all test cases doesn't exceed 2⋅1052⋅105.

Output

Output tt lines, where the ii-th line contains the number of good permutations in the ii-th test case modulo 109+7109+7.

Example

input

Copy

4
3
1 1 1
5
1 2 3 4 5
5
0 2 0 3 0
4
1 3 5 1

output

Copy

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 66 permutations possible with numbers from 11 to 33: [1,2,3][1,2,3], [1,3,2][1,3,2], [2,1,3][2,1,3], [2,3,1][2,3,1], [3,1,2][3,1,2], [3,2,1][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 3636 permutations for which the sequence is good. One of them is the permutation [1,5,4,2,3][1,5,4,2,3] which results in the sequence s=[0,0,3,2,0]s=[0,0,3,2,0]. This is a good sequence because

• s1=s2&s3&s4&s5=0s1=s2&s3&s4&s5=0,
• s1&s2=s3&s4&s5=0s1&s2=s3&s4&s5=0,
• s1&s2&s3=s4&s5=0s1&s2&s3=s4&s5=0,
• s1&s2&s3&s4=s5=0s1&s2&s3&s4=s5=0.
• ============================================================================================================================================
• 既然a1=a2&a3&a4&a5..&an

那么a1=a1&a2&a3....&an

也有an=a1&a2&a3...&an

所以a1和an的二进制pos位置如果是1，那么a1...anpos位置全是1，否则，必然有一个0

这样

a1&a2  ？ a3......&an 

单独看pos位置，是1当今当全是1，是0的话，左右两边的a1,an就完美的控制了永远是0，无论新加的数字是多少，这一位置都是0

方案数计算组合数即可，排列

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
using namespace std;
 
typedef long long int  ll;
map<int,ll>cnt;
 
# define mod 1000000007
 
ll fac[200000+10];
 
void init()
{
    fac[0]=1;
 
    for(int i=1; i<=200000; i++)
    {
        fac[i]=fac[i-1]*i%mod;
    }
}
 
ll a[2000000+10];
 
int main()
{
 
 
    init();
 
    int t;
 
    cin>>t;
 
    while(t--)
    {
        int n;
 
        cnt.clear();
        cin>>n;
        cin>>a[1];
        ll ans=a[1];
        cnt[a[1]]++;
        for(int i=2; i<=n; i++)
        {
            cin>>a[i];
            ans=(ans&a[i]);
            cnt[a[i]]++;
        }
 
        if(cnt[ans]>=2)
         cout<-1)%mod*fac[n-2]%mod<
         else
            cout<<0<
 
    }
    return 0;
}