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 = (long long)0x3f3f3f3f3f3f3f, MAX = 9e18, MIN = -9e18;
const int N = 1e6 + 10, M = 2e6 + 10, mod = 1e9 + 7, inf = 0x3f3f3f3f;
long long a[N];
vector<int> p1, p2;
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int _;
  _ = 1;
  while (_--) {
    long long n, ans = 0;
    cin >> n;
    for (int i = 1; (i) <= (n); i++) {
      cin >> a[i];
      if (a[i])
        p1.push_back(i);
      else
        p2.push_back(i);
    }
    int n1 = p1.size(), n2 = p2.size();
    for (int i = 0; (i) < (n1); i++) {
      vector<int>::iterator p = lower_bound(p2.begin(), p2.end(), p1[i]), q;
      if (p == p2.end()) {
        while (i < n1) {
          ans += abs(p1[i] - p2.back());
          p2.pop_back();
          i++;
        }
        break;
      }
      int j = p - p2.begin(), j2, f = 0;
      if (j > 0) {
        f = 1;
        j2 = j - 1;
        q = p;
        q--;
      }
      if (f &&
          (abs(p1[i] - p2[j2]) < abs(p1[i] - p2[j]) || (n1 - i) > (n2 - j))) {
        ans += abs(p1[i] - p2[j2]);
        p2.erase(q);
        n2--;
        continue;
      }
      ans += abs(p1[i] - p2[j]);
      p2.erase(p);
      n2--;
    }
    cout << ans << '\n';
  }
  return 0;
}
