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>
#define int long long
using namespace std;

const int N = 5e3 + 5;
int dp[N][N];
vector<int>o,z;

int fx(int i , int j){
   
   //My Bad : < 
   if(i == o.size()){
    return 0;
   }

   if(j == z.size()){
    if(i == o.size()){
        return 0;
    }
    return 1e18;
   }

   if(dp[i][j] != -1){
    return dp[i][j];
   }

   int a = abs(z[j] - o[i]) +  fx(i  + 1 , j + 1);
   int b = fx(i , j + 1);

   return dp[i][j] = min(a,b); 
}


signed main(){

    memset(dp , -1, sizeof(dp));
    int n ;
    cin >> n;
    
    int x;
    for(int i = 0 ; i < n ; i ++){
        cin >> x;
        if(x == 1){
            o.push_back(i);
        }else{
            z.push_back(i);
        }
    }

    if(o.size() == 0){
        cout<<0<<endl;
        return 0;
    }

    int ans = fx(0,0);
    cout<<ans<<endl;


    return 0;
}