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 int MOD = 1000000007;
int main() {
  int n;
  cin >> n;
  int ar1[n + 10], ar[n + 10];
  map<int, int> m;
  vector<int> v1, v2;
  for (int i = 1; i <= n; i++) {
    cin >> ar1[i];
  }
  int k = 1;
  for (int i = n; i >= 1; i--) {
    ar[k] = ar1[i];
    k++;
  }
  for (int i = 1; i <= n; i++) {
    if (ar[i]) {
      int mini = INT_MAX, ind;
      for (int j = i - 1; j >= 1; j--) {
        if ((ar[j] == 0) && (m[j] == 0)) {
          int dis = i - j;
          if (dis < mini) {
            mini = dis;
            ind = j;
          }
        }
      }
      for (int j = i + 1; j <= n; j++) {
        if ((ar[j] == 0) && (m[j] == 0)) {
          int dis = j - i;
          if (dis < mini) {
            mini = dis;
            ind = j;
          }
        }
      }
      m[ind]++;
      v1.push_back(i);
      v2.push_back(ind);
    }
  }
  sort(v1.begin(), v1.end());
  sort(v2.begin(), v2.end());
  long long int mini = INT_MAX, sum = 0;
  for (int i = 0; i < v1.size(); i++) sum += abs(v1[i] - v2[i]);
  mini = min(mini, sum);
  sum = 0;
  int j = v1.size() - 1;
  for (int i = 0; i < v1.size(); i++, j--) sum += abs(v1[i] - v2[j]);
  mini = min(mini, sum);
  cout << mini << endl;
  return 0;
}
