human
Give me a Python program to calculate the minimal number of swaps he would need to perform to make it so that the array does not contain two neighboring items with the same parity given the following:
PYTHON3 solution for "

William has an array of n integers a_1, a_2, ..., a_n. In one move he can swap two neighboring items. Two items a_i and a_j are considered neighboring if the condition |i - j| = 1 is satisfied.

William wants you to calculate the minimal number of swaps he would need to perform to make it so that the array does not contain two neighboring items with the same parity.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 10^4). Description of the test cases follows.

The first line of each test case contains an integer n (1 ≤ n ≤ 10^5) which is the total number of items in William's array.

The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) which are William's array.

It is guaranteed that the sum of n over all test cases does not exceed 10^5.

Output

For each test case output the minimal number of operations needed or -1 if it is impossible to get the array to a state when no neighboring numbers have the same parity.

Example

Input

5
3
6 6 1
1
9
6
1 1 1 2 2 2
2
8 6
6
6 2 3 4 5 1

Output

1
0
3
-1
2

Note

In the first test case the following sequence of operations would satisfy the requirements: 

  1. swap(2, 3). Array after performing the operation: [6, 1, 6] 


In the second test case the array initially does not contain two neighboring items of the same parity.

In the third test case the following sequence of operations would satisfy the requirements: 

  1. swap(3, 4). Array after performing the operation: [1, 1, 2, 1, 2, 2] 
  2. swap(2, 3). Array after performing the operation: [1, 2, 1, 1, 2, 2] 
  3. swap(4, 5). Array after performing the operation: [1, 2, 1, 2, 1, 2] 


In the fourth test case it is impossible to satisfy the requirements.

In the fifth test case the following sequence of operations would satisfy the requirements: 

  1. swap(2, 3). Array after performing the operation: [6, 3, 2, 4, 5, 1] 
  2. swap(4, 5). Array after performing the operation: [6, 3, 2, 5, 4, 1] ":
assistant
#Here is Python program to calculate the minimal number of swaps he would need to perform to make it so that the array does not contain two neighboring items with the same parity.
 #!/usr/bin/env python
import os
import sys
from io import BytesIO, IOBase
import math
from collections import Counter

def func(n, array):
    if n == 1:
        return 0
    array = [val % 2 for val in array]
    counter = Counter(array)
    if 0 not in counter or 1 not in counter:
        return -1
    if abs(counter[0] - counter[1]) > 1:
        return -1
    if counter[0] > counter[1]:
        start = 0
    elif counter[0] < counter[1]:
        start = 1
    else:
        start = None
    second = False
    if start is None:
        start = array[0]
        second = True
    count = 0
    index = 0
    for i, val in enumerate(array):
        if val == start:
            count += abs(i - index)
            index += 2

    desire = sum(list(range(0, n, 2)))
    if not second:
        return count
    count2 = 0
    index = 1
    for i, val in enumerate(array):
        if val == start:
            count2 += abs(i - index)
            index += 2
    return min(count, count2)

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,...