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 gcd(int a, int b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
int shif(int a1[], int n, vector<int> v1, int j) {}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  long long a1[n];
  vector<int> v1;
  for (int i = 0; i < n; i++) {
    cin >> a1[i];
    if (a1[i]) v1.push_back(i);
  }
  long long sum = 0;
  if (a1[0] == 1) {
    long long j = 1;
    while (a1[j] != 0) j++;
    a1[j] = 2;
    a1[0] = INT_MAX;
    sum += j;
  }
  if (a1[n - 1] == 1) {
    long long j = n - 2;
    while (a1[j] != 0) j--;
    a1[j] = 2;
    a1[n - 1] = INT_MAX;
    sum += n - 1 - j;
  }
  long long i = 1;
  long long k = n - 2;
  while (i <= k) {
    if (a1[i] == 1) {
      long long s1 = INT_MAX;
      long long j1 = -1;
      for (int j = i - 1; j >= 0; j--) {
        if (a1[j] == 0) {
          a1[j] = 2;
          a1[i] = INT_MAX;
          s1 = abs(j - i);
          j1 = j;
          break;
        }
      }
      long long s2 = INT_MAX;
      long long j2 = -1;
      for (int j = i + 1; j < n; j++) {
        if (a1[j] == 0) {
          a1[j] = 2;
          a1[i] = INT_MAX;
          s2 = abs(j - i);
          j2 = j;
          break;
        }
      }
      if (s1 <= s2) {
        sum += s1;
        a1[j2] = 0;
      } else {
        sum += s2;
        a1[j1] = 0;
      }
    }
    if (a1[k] == 1) {
      long long s1 = INT_MAX;
      long long j1 = -1;
      for (int j = k - 1; j >= 0; j--) {
        if (a1[j] == 0) {
          a1[j] = 2;
          a1[k] = INT_MAX;
          s1 = abs(j - k);
          j1 = j;
          break;
        }
      }
      long long s2 = INT_MAX;
      long long j2 = -1;
      for (int j = k + 1; j < n; j++) {
        if (a1[j] == 0) {
          a1[j] = 2;
          a1[k] = INT_MAX;
          s2 = abs(j - k);
          j2 = j;
          break;
        }
      }
      if (s1 < s2) {
        sum += s1;
        a1[j2] = 0;
      } else {
        sum += s2;
        a1[j1] = 0;
      }
    }
    i++;
    k--;
  }
  cout << sum << "\n";
}
