human
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. 
Using python3 can you solve the prior task?

Can you provide a Python code to find the number of permutations p of numbers ranging from 1 to n, for which the sequence a_{p_1}, a_{p_2},?
assistant
#Here is Python code to find the number of permutations p of numbers ranging from 1 to n, for which the sequence a_{p_1}, a_{p_2}, .

import os
import sys
from io import BytesIO, IOBase

def main():
    import bisect
    import math
    # import itertools
    # import heapq
    # from queue import PriorityQueue, LifoQueue, SimpleQueue

    # import sys.stdout.flush() use for interactive problems
    alpha = 'abcdefghijklmnopqrstuvwxyz'
    ALPHA = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
    inf = 1e17
    mod = 10 ** 9 + 7

    # Max = 10**6
    # primes = []
    # prime = [True for i in range(10**6+1)]
    # p = 2
    # while (p * p <= Max+1):
    #
    #     # If prime[p] is not
    #     # changed, then it is a prime
    #     if (prime[p] == True):
    #
    #         # Update all multiples of p
    #         for i in range(p * p, Max+1, p):
    #             prime[i] = False
    #     p += 1
    #
    # for p in range(2, Max+1):
    #     if prime[p]:
    #         primes.append(p)

    def factorial(n):
        M = 1000000007
        f = 1

        for i in range(1, n + 1):
            f = (f * i) % M # Now f never can
        # exceed 10^9+7

        return f

    def ncr(n, r, p):
        # initialize numerator
        # and denominator
        num = den = 1
        for i in range(r):
            num = (num * (n - i)) % p
            den = (den * (i + 1)) % p
        return (num * pow(den,
                          p - 2, p)) % p




    def solve(n,arr):
        Min = min(arr)
        cnt = arr.count(Min)
        
        for ele in arr:
            if ele & Min != Min:
                return 0
        if cnt >= 0:
            const = factorial(n-2)
            answer = ((cnt % mod) * ((cnt-1)% mod)*const)% mod
            return answer%mod
        return 0







    t = int(input())
    ans = []
    for _ in range(t):
        n = int(input())
        #n,k = map(int, input().split())
        arr = [int(x) for x in input().split()]
        # arr = list(input())
        # s = input()
        # t = input()
        # grid = []
        # for i in range(n):
        #     grid.append(list(input()))

        ans.append(solve(n,arr))

    for...