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>

typedef long long ll;

const int MAX_SIZE = 5e3 + 1;
ll persons[MAX_SIZE], chairs[MAX_SIZE];
ll dp[MAX_SIZE][MAX_SIZE];

int main()
{
    ll n {0}, total_persons {0}, total_chairs {0};
    ll occupied {1}, empty {0}, num;
    std::cin >> n;
    for (ll i = 0; i < n; i++) {
        std::cin >> num;
        if (num == occupied) {
            persons[++total_persons] = i;
        }
        else {
            chairs[++total_chairs] = i;
        }
    }
    for (int p = 1; p <= total_persons; p++) {
        for (int c = p; c <= total_chairs; c++) {
            dp[p][c] = dp[p - 1][c - 1] + std::abs(persons[p] - chairs[c]);
            if (c - 1 >= p) {
                dp[p][c] = std::min(dp[p][c],
                                    dp[p][c - 1]);
            }
        }
    }
    std::cout << dp[total_persons][total_chairs] << std::endl;

    return 0;
}