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 <iostream>
#include <cstring>
#include <algorithm>
#include <vector>

using namespace std;

const int MAX_N = 5050;
const int INF = (int)1e9;

int n;
int a[MAX_N];
int dp[MAX_N][MAX_N];
int mn[MAX_N][MAX_N];
vector<int> people, seats;

int main() {
  for (int i = 0; i < MAX_N; ++ i) {
    for (int j = 0; j < MAX_N; ++ j) {
        dp[i][j] = INF;
        mn[i][j] = INF;
    }
  }
  cin >> n;
  for (int i = 0; i < n; ++ i) {
    cin >> a[i];
    if (a[i]) {
      people.push_back(i);
    } else {
      seats.push_back(i);
    }
  }
  if (people.size() == 0) {
    cout << 0 << endl;
    return 0;
  }
  int ans = INF;
  for (int i = 0; i < seats.size(); ++ i) {
    dp[0][i] = abs(people[0] - seats[i]);
    if (i > 0) {
      mn[0][i] = min(mn[0][i - 1], dp[0][i]);
    } else {
      mn[0][i] = dp[0][i];
    }
  }
  for (int i = 1; i < people.size(); ++ i) {
    for (int j = 1; j < seats.size(); ++ j) {
      dp[i][j] = min(dp[i][j], mn[i - 1][j - 1] + abs(people[i] - seats[j]));
      mn[i][j] = min(mn[i][j - 1], dp[i][j]);
    }
  }
  for (int i = 0; i < seats.size(); ++ i) {
    ans = min(ans, dp[people.size() - 1][i]);
  }
  cout << ans << endl;
  return 0;
}
