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 int A[1001];
int32_t main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long int t = 1;
  while (t--) {
    long long int n;
    cin >> n;
    long long int arr[n];
    set<long long int> s, s1;
    for (long long int i = 0; i < n; i++) {
      cin >> arr[i];
      if (arr[i]) {
        s.insert(i);
      } else {
        s1.insert(i);
      }
    }
    long long int ans = 0;
    for (auto i : s) {
      auto it1 = s1.upper_bound(i);
      if (it1 == s1.begin()) {
        ans += abs(i - *it1);
        s1.erase(it1);
      } else if (it1 == s1.end()) {
        it1--;
        ans += abs(i - *it1);
        s1.erase(it1);
      } else {
        long long int x = *it1;
        auto it2 = it1;
        it1--;
        long long int x2 = *it1;
        if (abs(x - i) < abs(x2 - i)) {
          ans += abs(x - i);
          s1.erase(it2);
        } else {
          ans += abs(x2 - i);
          s1.erase(it1);
        }
      }
    }
    cout << ans << endl;
  }
}
