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 <sstream>
#include <algorithm>
#include <vector>
#include <tuple>
#include <deque>
#include <queue>
#include <set>
#include <map>
#include <regex>
#include <unordered_set>
#include <unordered_map>
#include <climits>
#include <iomanip>
#include <math.h>
// #include "graphs.h"

#define ll long long
#define ld long double
#define INFL LLONG_MAX
#define INFI INT_MAX
#define PI 3.14159265358979323846
#define E 2.7182818284
#define MOD 1000000007

using namespace std;

int main() {

    ios::sync_with_stdio(0);
    cin.tie(0);

//    std::cout << std::fixed;
//    std::cout << std::setprecision(8);

//    int t;
//    cin >> t;
//
//    for (int ti = 1; ti <= t; ti++) {
//        cout << "Case #" << ti << ": ";
        int n;
        cin >> n;

        vector<int> starters;
        int a[n];
        for (int i = 0; i < n; i++) {
            cin >> a[i];
            if (a[i] == 1) starters.push_back(i);
        }

        int dp[n+1][n+1];
        for (int i = 0; i <= n; i++)
            for (int j = 0; j <= n; j++)
                dp[i][j] = 10000000;

        dp[0][0] = 0;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (a[i] == 1) {
                    dp[i+1][j] = min(dp[i+1][j], dp[i][j]);
                    continue;
                }
                if (j >= starters.size()) {
                    dp[i+1][j] = min(dp[i+1][j], dp[i][j]);
                    continue;
                }
                dp[i+1][j] = min(dp[i+1][j], dp[i][j]);
                int distance = abs(starters[j] - i);
                dp[i+1][j+1] = min(dp[i+1][j+1], dp[i][j] + distance);
            }
        }

        int ans = dp[n][starters.size()];
        cout << ans << endl;


//    }
}

/**
 NOOB CHECK

 - Integers overflow
 - Initialize arrays
 - Don't continue to next test case before finishing reading the input
 - Add 2 lines to speed up the input
*/