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 long long N = 5015;
long long n, laso = 0;
long long a[N], beh[N];
bool vis[N];
signed main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    vis[i] = a[i];
    if (a[i] == 1) laso = max(laso, i);
  }
  for (long long i = n; i >= 1; i--) beh[i] = beh[i + 1] + (a[i] == 1);
  long long ans = 0;
  for (long long i = 1; i <= n; i++) {
    if (a[i] == 0) continue;
    if (i == laso) {
      long long mins = 0x3f3f3f3f, pos = 0;
      for (long long j = 1; j <= n; j++) {
        if (i == j || vis[j] == 1 || a[j] == 1) continue;
        if (abs(i - j) < mins) {
          mins = abs(i - j);
          pos = j;
        }
      }
      ans += mins;
      break;
    } else {
      if (a[i + 1] == 1) {
        long long mins = 0x3f3f3f3f, pos = 0;
        for (long long j = 1; j < i; j++) {
          if (i == j || vis[j] == 1 || a[j] == 1) continue;
          if (abs(i - j) < mins) {
            mins = abs(i - j);
            pos = j;
          }
          if (mins == abs(i - j)) pos = max(pos, j);
        }
        if (!pos) {
          for (long long j = i + 1; j <= n; j++) {
            if (vis[j] == 1 || a[j] == 1) continue;
            if (abs(i - j) < mins) {
              mins = abs(i - j);
              pos = j;
            }
            if (mins == abs(i - j)) pos = min(pos, j);
          }
        }
        ans += mins;
        vis[pos] = 1;
      } else if (a[i + 1] == 0) {
        long long num = 0;
        for (long long j = i + 1; j <= n; j++) {
          if (a[j] == 1) break;
          if (vis[j] == 1) continue;
          num++;
        }
        if (num > beh[i + 1]) {
          for (long long j = i + 1; j <= n; j++) {
            if (a[j] == 1) break;
            if (vis[j] == 1) continue;
            ans += abs(i - j);
            vis[j] = 1;
            break;
          }
        } else {
          long long mins = 0x3f3f3f3f, pos = 0;
          for (long long j = 1; j < i; j++) {
            if (i == j || vis[j] == 1 || a[j] == 1) continue;
            if (abs(i - j) < mins) {
              mins = abs(i - j);
              pos = j;
            }
            if (mins == abs(i - j)) pos = max(pos, j);
          }
          if (!pos) {
            for (long long j = i + 1; j <= n; j++) {
              if (vis[j] == 1 || a[j] == 1) continue;
              if (abs(i - j) < mins) {
                mins = abs(i - j);
                pos = j;
              }
              if (mins == abs(i - j)) pos = min(pos, j);
            }
          }
          ans += mins;
          vis[pos] = 1;
        }
      }
    }
  }
  printf("%lld\n", ans);
  return 0;
}
