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 = 1000000007;
long long modpow(long long x, long long n, long long m) {
  if (n == 0) return 1 % m;
  long long u = modpow(x, n / 2, m);
  u = (u * u) % m;
  if (n % 2 == 1) u = (u * x) % m;
  return u;
}
void solve() {
  long long n, ans = 0, o = 0;
  cin >> n;
  set<long long> se;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i] == 0)
      se.insert(i);
    else
      o++;
  }
  for (long long i = 0; i < n; i++) {
    if (v[i] == 1) {
      auto it = se.lower_bound(i);
      if (it == se.begin()) {
        long long pos = *it;
        se.erase(pos);
        ans += abs(pos - i);
      } else if (it == se.end()) {
        it--;
        long long pos = *it;
        se.erase(pos);
        ans += abs(pos - i);
      } else {
        if (o == se.size()) {
          it--;
          long long lp = *it;
          se.erase(lp);
          ans += abs(i - lp);
        } else {
          long long r = *it;
          it--;
          long long l = *it;
          if (abs(i - l) < abs(i - r)) {
            se.erase(l);
            ans += abs(i - l);
          } else {
            se.erase(r);
            ans += abs(i - r);
          }
        }
      }
      o--;
    }
  }
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t = 1;
  while (t--) solve();
}
