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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  int arr[n + 1];
  for (int i = 1; i <= n; i++) cin >> arr[i];
  long long int ans = 0;
  for (int i = 1; i < n; i++) {
    int x = INT_MAX;
    int y = INT_MAX;
    int x1, y1;
    if (arr[i] == 1) {
      for (int j = i + 1; j <= n; j++) {
        if (arr[j] == 0) {
          x = abs(i - j);
          x1 = j;
          break;
        }
      }
      if (i >= 2) {
        for (int j = i - 1; j >= 1; j--) {
          if (arr[j] == 0) {
            y = abs(i - j);
            y1 = j;
            break;
          }
        }
      }
      if (x < y) {
        ans += x;
        arr[x1] = 2;
      } else {
        ans += y;
        arr[y1] = 2;
      }
    }
  }
  if (arr[n] == 1) {
    for (int j = n - 1; j >= 1; j--) {
      if (arr[j] == 0) {
        ans += abs(n - j);
        arr[j] = 2;
        break;
      }
    }
  }
  cout << ans << endl;
}
