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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int i, j, k, l;
  long long x, y, z;
  long long n;
  cin >> n;
  vector<int> a(n);
  vector<int> b(n);
  int count = 0;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
    if (a[i] == 1) count++;
  }
  vector<int> pre(n + 1, -1);
  vector<int> post(n + 1, -1);
  for (i = 0; i < n; i++) {
    if (b[i] == 1) {
      j = i - 1;
      while (j >= 0) {
        if (b[j] == 0) {
          b[j] = 2;
          pre[i] = i - j;
          break;
        } else
          j--;
      }
      if (pre[i] == -1) {
        j = i + 1;
        while (j < n) {
          if (b[j] == 0) {
            b[j] = 3;
            pre[i] = j - i;
            break;
          } else
            j++;
        }
      }
    }
  }
  for (i = 0; i < n; i++) {
    if (b[i] != 3) b[i] = a[i];
  }
  for (i = n - 1; i >= 0; i--) {
    if (b[i] == 1) {
      if (b[pre[i]] == 3)
        post[i] = pre[i];
      else {
        j = i + 1;
        while (j < n) {
          if (b[j] == 0) {
            b[j] = 2;
            post[i] = j - i;
            break;
          } else
            j++;
        }
        if (post[i] == -1) {
          j = i - 1;
          while (j >= 0) {
            if (b[j] == 0) {
              b[j] = 2;
              post[i] = i - j;
              break;
            } else
              j--;
          }
        }
      }
    }
  }
  vector<int> pk;
  for (i = 0; i < n; i++) {
    if (a[i] == 1) pk.push_back(i);
  }
  int ans = 0, mina = 0;
  for (i = 0; i < pk.size(); i++) {
    ans += pre[pk[i]];
  }
  mina = ans;
  for (i = pk.size() - 1; i >= 0; i--) {
    ans += (post[pk[i]] - pre[pk[i]]);
    mina = min(ans, mina);
  }
  cout << mina;
  return 0;
}
