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 long long inf = 1e18;
const long long N = 200005;
long long n;
long long getans(vector<long long> a) {
  stack<long long> st, en, start;
  long long val = 0;
  for (long long i = n - 1; i >= 0; i--)
    if (a[i] == 0) en.push(i);
  for (long long i = 0; i < n; i++) {
    if (a[i] == 1) {
      if (start.empty()) {
        val += (abs(i - en.top()));
        en.pop();
      } else {
        if ((!en.empty()) && abs(i - en.top()) < abs(i - start.top())) {
          val += (abs(i - en.top()));
          en.pop();
        } else {
          val += (abs(i - start.top()));
          start.pop();
        }
      }
    } else {
      if (!en.empty() && i == en.top()) {
        start.push(en.top());
        en.pop();
      }
    }
  }
  return val;
}
void solve() {
  long long i, j, k;
  cin >> n;
  vector<long long> a(n), b;
  for (long long i = 0; i < n; i++) cin >> a[i];
  b = a;
  reverse(a.begin(), a.end());
  cout << min(getans(a), getans(b)) << "\n";
  return;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long t = 1, i, j, k;
  while (t--) {
    solve();
  }
  return 0;
}
