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. 
#!/us
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}, .
r/bin/env python
import os
import sys
from io import BytesIO, IOBase
import math
from collections import Counter

def func(n, array):
    curr_and = array[0]
    min_val = array[0]
    count_min = 1
    for i in array[1:]:
        if i < min_val:
            min_val = i
            count_min = 1
        elif i == min_val:
            count_min += 1
        curr_and = curr_and & i
    if curr_and < min_val:
        return 0
    return count_min * (count_min - 1) * math.factorial(n - 2) % (10 ** 9 + 7)

def main():
    num_test = int(parse_input())
    result = []
    for _ in range(num_test):
        n = int(parse_input())
        array = [int(i) for i in parse_input().split()]
        result.append(func(n, array))
    print("\n".join(map(str, result)))

# region fastio

# BUFSIZE = 8192

# class FastIO(IOBase):
#     newlines = 0

#     def __init__(self, file):
#         self._fd = file.fileno()
#         self.buffer = BytesIO()
#         self.writable = "x" in file.mode or "r" not in file.mode
#         self.write = self.buffer.write if self.writable else None

#     def read(self):
#         while True:
#             b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
#             if not b:
#                 break
#             ptr = self.buffer.tell()
#             self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
#         self.newlines = 0
#         return self.buffer.read()

#     def readline(self):
#         while self.newlines == 0:
#             b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
#             self.newlines = b.count(b"\n") + (not b)
#             ptr = self.buffer.tell()
#             self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
#         self.newlines -= 1
#         return self.buffer.readline()

#     def flush(self):
#         if self.writable:
#             os.write(self._fd, self.buffer.getvalue())
#             self.buffer.truncate(0), self.buffer.seek(0)

# class IOWrapper(IOBase):
#     def...