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;
void fast() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int32_t main() {
  fast();
  long long n;
  cin >> n;
  long long arr[n], brr[n];
  for (long long i = 0; i < n; i++) {
    cin >> arr[i];
    brr[i] = 0;
  }
  long long ans = 0, sum = 0;
  for (long long i = 0; i < n; i++) {
    if (arr[i] == 1) {
      long long idx1 = -1, idx2 = -1;
      for (long long j = i + 1; j < n; j++) {
        if (arr[j] == 0) {
          idx1 = j;
          break;
        }
      }
      for (long long j = i - 1; j >= 0; j--) {
        if (arr[j] == 0) {
          idx2 = j;
          break;
        }
      }
      if (idx1 == -1) {
        sum += i - idx2;
        arr[idx2] = -1;
        brr[idx2] = 1;
      } else if (idx2 == -1) {
        sum += idx1 - i;
        arr[idx1] = -1;
        brr[idx1] = 1;
      } else {
        if (i - idx2 <= idx1 - i) {
          sum += i - idx2;
          arr[idx2] = -1;
          brr[idx2] = 1;
        } else {
          sum += idx1 - i;
          arr[idx1] = -1;
          brr[idx1] = 1;
        }
      }
    }
  }
  for (long long i = 0; i < n; i++) {
    if (brr[i] == 1) {
      long long idx1 = -1, idx2 = -1;
      for (long long j = i + 1; j < n; j++) {
        if (arr[j] == 1) {
          idx1 = j;
          break;
        }
      }
      for (long long j = i - 1; j >= 0; j--) {
        if (arr[j] == 1) {
          idx2 = j;
          break;
        }
      }
      if (idx1 == -1) {
        ans += i - idx2;
        arr[idx2] = -1;
      } else if (idx2 == -1) {
        ans += idx1 - i;
        arr[idx1] = -1;
      } else {
        if (i - idx2 <= idx1 - i) {
          ans += i - idx2;
          arr[idx2] = -1;
        } else {
          ans += idx1 - i;
          arr[idx1] = -1;
        }
      }
    }
  }
  cout << ans << "\n";
  return 0;
}
