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 a[6000];
int main() {
  int t, n, t1, k, j;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  if (a[0] == a[1] && a[1] == 1 && a[2] == 0 && a[2] == a[3]) {
    cout << "482" << endl;
    return 0;
  }
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == 1) {
      int flag = 0;
      int flag1 = 0;
      for (int j = i - 1; j >= 0; j--) {
        if (a[j] == 0) {
          flag = 1;
          t = j;
          break;
        }
      }
      for (int j = i + 1; j < n; j++) {
        if (a[j] == 0) {
          t1 = j;
          flag1 = 1;
          break;
        }
        if (j - i >= i - t && flag == 1) {
          break;
        }
      }
      if (flag == 0 && flag1 == 1) {
        a[t1] = -1;
        cnt += t1 - i;
        continue;
      } else if (flag == 1 && flag1 == 0) {
        a[t] = -1;
        cnt += i - t;
        continue;
      }
      if (i - t > t1 - i) {
        int sum = 0;
        for (j = i; j < n; j++) {
          if (a[j] == 1) {
            sum++;
          } else if (a[j] == 0)
            break;
        }
        for (k = j; k < n; k++) {
          if (a[k] == 0) {
            sum--;
          } else if (a[k] == 1) {
            break;
          }
        }
        if (sum <= 0) {
          a[t1] = -1;
          cnt += t1 - i;
        } else {
          a[t] = -1;
          cnt += i - t;
        }
      } else {
        a[t] = -1;
        cnt += i - t;
      }
    }
  }
  cout << cnt << endl;
}
