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 int N = 50005;
int a[N], n;
int b[N], ans;
set<int> s;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i)
    if (!a[i]) s.insert(i);
  for (int i = n - 1; i >= 1; --i, b[i] = b[i + 1])
    if (a[i + 1])
      --b[i];
    else
      ++b[i];
  for (int i = 1; i <= n; ++i) {
    if (!a[i]) continue;
    set<int>::iterator it;
    if (b[i] > 0 || i == n) {
      it = s.lower_bound(i);
      int p1 = 0, p2 = 0;
      if (it != s.begin()) p1 = *(--it), ++it;
      if (it != s.end()) p2 = *it;
      if (!p1)
        ans += p2 - i, s.erase(p2);
      else if (!p2)
        ans += i - p1, s.erase(p1);
      else {
        if (p2 - i < i - p1)
          s.erase(p2);
        else
          s.erase(p1);
        ans += min(i - p1, p2 - i);
      }
    } else {
      it = s.lower_bound(i), --it;
      ans += i - *it, s.erase(*it);
    }
  }
  printf("%d", ans);
  return 0;
}
