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 apSum(long long int a, long long int n, long long int d = 1) {
  long long int res = (2 * a + (n - 1) * d) * n;
  res /= 2;
  return res;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool isPrime(long long int n) {
  for (long long int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> list(n), high, low;
  for (long long int i = 0; i < n; i++) {
    cin >> list[i];
    if (list[i] == 1)
      high.push_back(i);
    else
      low.push_back(i);
  }
  long long int ans = 0;
  vector<bool> mark(n + 1, false);
  for (long long int i = 0; i < high.size(); i++) {
    long long int v =
        upper_bound(low.begin(), low.end(), high[i]) - low.begin();
    if (v == 0) v += 1;
    if (!mark[v - 1]) {
      mark[v - 1] = true;
      ans += abs(high[i] - low[v - 1]);
      continue;
    }
    long long int first = v - 2;
    long long int last = v;
    while (true) {
      if (first < 0 && last >= low.size()) break;
      long long int res = 0;
      if (first >= 0 && !mark[first]) {
        ans += abs(high[i] - low[first]);
        mark[first] = true;
        break;
      }
      if (last < n && !mark[last]) {
        ans += abs(high[i] - low[last]);
        mark[last] = true;
        break;
      }
      first -= 1;
      last += 1;
    }
  }
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int testCases = 1;
  while (testCases--) {
    solve();
  }
}
