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 i, j, k, n, m, a[5050], res, l, r, res2, res3, res4;
vector<int> v, v2;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (a[i]) {
      v.push_back(i);
    }
  }
  for (i = 1; i <= n; i++) {
    if (a[i] == 1) {
      l = r = 0;
      for (j = i; j >= 1; j--) {
        if (!a[j]) {
          l = j;
          break;
        }
      }
      for (j = i; j <= n; j++) {
        if (!a[j]) {
          r = j;
          break;
        }
      }
      if (!l) {
        a[r] = 2;
        v2.push_back(r);
        continue;
      }
      if (!r) {
        a[l] = 2;
        v2.push_back(l);
        continue;
      }
      if (i - l <= r - i) {
        a[l] = 2;
        v2.push_back(l);
      } else {
        a[r] = 2;
        v2.push_back(r);
      }
    }
  }
  sort(v2.begin(), v2.end());
  for (i = 0; i < v.size(); i++) {
    res += abs(v[i] - v2[i]);
  }
  for (i = 1; i <= n; i++) {
    if (a[i] == 2) {
      a[i] = 0;
    }
  }
  v2.clear();
  for (i = 1; i <= n; i++) {
    if (a[i] == 1) {
      l = r = 0;
      for (j = i; j >= 1; j--) {
        if (!a[j]) {
          l = j;
          break;
        }
      }
      for (j = i; j <= n; j++) {
        if (!a[j]) {
          r = j;
          break;
        }
      }
      if (!l) {
        a[r] = 2;
        v2.push_back(r);
        continue;
      }
      if (!r) {
        a[l] = 2;
        v2.push_back(l);
        continue;
      }
      if (i - l < r - i) {
        a[l] = 2;
        v2.push_back(l);
      } else {
        a[r] = 2;
        v2.push_back(r);
      }
    }
  }
  sort(v2.begin(), v2.end());
  for (i = 0; i < v.size(); i++) {
    res2 += abs(v[i] - v2[i]);
  }
  for (i = 1; i <= n; i++) {
    if (a[i] == 2) {
      a[i] = 0;
    }
  }
  printf("%d", min(res, res2));
}
