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;
long long binpow(long long a, long long b) {
  long long exp = 1000000007;
  a %= exp;
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = ((res) * (a)) % exp;
    a = ((a) * (a)) % exp;
    b >>= 1;
  }
  return res;
}
bool isprime(long long x) {
  for (long long d = 2; d * d <= x; d++) {
    if (x % d == 0) return false;
  }
  return true;
}
unsigned int cntsb(unsigned int n) {
  unsigned int count = 0;
  while (n) {
    count += n & 1;
    n >>= 1;
  }
  return count;
}
int LCSubStr(string X, string Y, int m, int n) {
  int LCSuff[m + 1][n + 1];
  int result = 0;
  for (int i = 0; i <= m; i++) {
    for (int j = 0; j <= n; j++) {
      if (i == 0 || j == 0)
        LCSuff[i][j] = 0;
      else if (X[i - 1] == Y[j - 1]) {
        LCSuff[i][j] = LCSuff[i - 1][j - 1] + 1;
        result = max(result, LCSuff[i][j]);
      } else
        LCSuff[i][j] = 0;
    }
  }
  return result;
}
void solve(int cas) {}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int number = 0;
  int n;
  cin >> n;
  int a[n];
  for (int i = 0; i < n; i++) cin >> a[i];
  ;
  int vis[n];
  for (int i = 0; i < n; i++) {
    vis[i] = a[i];
  }
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    if (a[i] == 1) {
      int left = i - 1, right = i + 1;
      bool fr = 0, fl = 0;
      int pos1 = 500000, pos2 = 5000000;
      while (left >= 0) {
        if (a[left] == 0 && vis[left] == 0) {
          pos1 = left;
          fl = 1;
          break;
        }
        left--;
      }
      while (right < n) {
        if (a[right] == 0 && vis[right] == 0) {
          pos2 = right;
          fr = 1;
          break;
        }
        right++;
      }
      if (fl == 0 && fr == 1) {
        ans += (pos2 - i);
        vis[pos2] = 1;
      } else if (fl == 1 && fr == 0) {
        ans += (i - pos1);
        vis[pos1] = 1;
      } else {
        if (pos2 - i < i - pos1) {
          ans += (pos2 - i);
          vis[pos2] = 1;
        } else {
          ans += (i - pos1);
          vis[pos1] = 1;
        }
      }
    }
  }
  cout << ans;
  return 0;
}
