There are n armchairs, numbered from 1 to n from left to right. Some armchairs are occupied by people (at most one person per armchair), others are not. The number of occupied armchairs is not greater than n/2.

For some reason, you would like to tell people to move from their armchairs to some other ones. If the i-th armchair is occupied by someone and the j-th armchair is not, you can tell the person sitting in the i-th armchair to move to the j-th armchair. The time it takes a person to move from the i-th armchair to the j-th one is |i - j| minutes. You may perform this operation any number of times, but these operations must be done sequentially, i. e. you cannot tell a person to move until the person you asked to move in the last operation has finished moving to their destination armchair.

You want to achieve the following situation: every seat that was initially occupied must be free. What is the minimum time you need to do it?

Input

The first line contains one integer n (2 ≤ n ≤ 5000) — the number of armchairs.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1). a_i = 1 means that the i-th armchair is initially occupied, a_i = 0 means that it is initially free. The number of occupied armchairs is at most n/2.

Output

Print one integer — the minimum number of minutes you have to spend to achieve the following situation: every seat that was initially occupied must be free.

Examples

Input


7
1 0 0 1 0 0 1


Output


3


Input


6
1 1 1 0 0 0


Output


9


Input


5
0 0 0 0 0


Output


0

Note

In the first test, you can perform the following sequence:

  1. ask a person to move from armchair 1 to armchair 2, it takes 1 minute; 
  2. ask a person to move from armchair 7 to armchair 6, it takes 1 minute; 
  3. ask a person to move from armchair 4 to armchair 5, it takes 1 minute. 



In the second test, you can perform the following sequence:

  1. ask a person to move from armchair 1 to armchair 4, it takes 3 minutes; 
  2. ask a person to move from armchair 2 to armchair 6, it takes 4 minutes; 
  3. ask a person to move from armchair 4 to armchair 5, it takes 1 minute; 
  4. ask a person to move from armchair 3 to armchair 4, it takes 1 minute. 



In the third test, no seat is occupied so your goal is achieved instantly.
#include <bits/stdc++.h>
using namespace std;

const int INF = 1e9;

int p[5010], q[5010];
int dp[5010][5010];
int n;

int main()
{
	scanf("%d", &n);
	int lenp = 0, lenq = 0;
	for(int i=1; i<=n; i++) {
		int tmp;
		scanf("%d", &tmp);
		if(tmp == 0) p[++lenp] = i;
		else q[++lenq] = i;
	}
	//初始化 
	for(int i=0; i<=lenp; i++) {
		for(int j=0; j<=lenq; j++) {
			dp[i][j] = INF;
		}
	}
	for(int i=0; i<=lenp; i++) dp[i][0] = 0;
	dp[1][1] = abs(p[1] - q[1]);
	//dp
	for(int i=1; i<lenp; i++) {
		for(int j=1; j<=i+1; j++) {
			if(j > lenq) break;
			dp[i+1][j] = min(dp[i][j], dp[i][j-1] + abs(p[i+1] - q[j]));
		}
	}
	
	printf("%d\n", dp[lenp][lenq]);
	return 0;
}