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 mod = 1e9 + 7;
const long long mx = 2e6 + 10;
int posx[] = {1, -1, 0, 0};
int posy[] = {0, 0, 1, -1};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1, n, k, m, a, b, c, d;
  while (t--) {
    cin >> n;
    vector<long long> v(n);
    for (auto &i : v) cin >> i;
    set<long long> free;
    for (int i = 0; i < n; i++) {
      if (v[i] == 0) free.insert(i);
    }
    long long ans = 0;
    for (int i = 0; i < n; i++) {
      if (v[i] == 0) continue;
      auto it = free.lower_bound(i);
      if (it == free.begin()) {
        ans += abs(i - *it);
        free.erase(it);
      } else if (it == free.begin()) {
        it--;
        ans += abs(i - *it);
        free.erase(*it);
      } else {
        it--;
        auto it2 = it;
        it2++;
        long long value1 = *it2;
        long long value2 = *it;
        if (abs(*it - i) > abs(*it2 - i)) {
          ans += abs(value1 - i);
          free.erase(it2);
        } else {
          ans += abs(value2 - i);
          free.erase(it);
        }
      }
    }
    cout << ans << '\n';
  }
  return 0;
}
