description: <image>
  
  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] 
solution: import java.io.*;
  import java.util.*;
   
  public class Codeforces {
  	public static void main(String[] args) throws IOException {
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		int cases = Integer.parseInt(br.readLine());
  		o:while(cases-- > 0) {
  			String[] str = br.readLine().split(" ");
  			int n = Integer.parseInt(str[0]);
  			int[] a = new int[n];
  			ArrayList<Integer> even = new ArrayList<>();
  			ArrayList<Integer> odd = new ArrayList<>();
  			str = br.readLine().split(" ");
  			for(int i=0; i<n; i++) {
  				a[i] = Integer.parseInt(str[i]);
  				if(a[i] % 2 != 0) {
  					odd.add(i);
  				}else {
  					even.add(i);
  				}
  			}
  			//System.out.println(even);
  			//System.out.println(odd);
  			if(n % 2 == 0) {
  				if(odd.size() != even.size()) {
  					System.out.println(-1);
  					continue;
  				}
  				int pe=0, po=0;
  				boolean iseven = true;
  				long count = 0;
  				for(int i=0; i<n; i++) {
  					if(iseven) {
  						count += Math.max(0, even.get(pe++) - i);
  					}else {
  						count += Math.max(0, odd.get(po++) - i);
  					}
  					//System.out.println(count);
  					iseven = !iseven;
  				}
  				long count2 = 0; pe=0; po=0;
  				iseven = false;
  				//System.out.println("====");
  				for(int i=0; i<n; i++) {
  					if(iseven) {
  						count2 += Math.max(0, even.get(pe++) - i);
  					}else {
  						count2 += Math.max(0, odd.get(po++) - i);
  					}
  					//System.out.println(count2);
  					iseven = !iseven;
  				}
  				System.out.println(Math.min(count, count2));
  			}else {
  				if(Math.abs(odd.size() - even.size()) != 1) {
  					System.out.println(-1);
  					continue;
  				}
  				boolean iseven = false;
  				if(odd.size() < even.size()) {
  					iseven = true;
  				}
  				int pe=0, po=0;
  				long count = 0;
  				for(int i=0; i<n; i++) {
  					if(iseven) {
  						count += Math.max(0, even.get(pe++) - i);
  					}else {
  						count += Math.max(0, odd.get(po++) - i);
  					}
  					iseven = !iseven;
  				}
  				System.out.println(count);
  			}
  			
  		}
  	}
  	public static int gcd(int a, int b) {
  		if(b == 0) {
  			return a;
  		}
  		return gcd(b, a%b);
  	}
  }