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 <cstring>
#include <vector>
#include <climits>
#include <bits/stdc++.h>
using namespace std;
vector<long long int >v,v1;
long long int d[5001][5001];
long long int slove(long long int i, long long int j)
{
    if(i>=v.size())
    {
        return 0;
    }
    if(j>=v1.size())
    {
        return 1e15;
    }

    if(d[i][j]!=-1)
        return d[i][j];
    d[i][j]=min(abs(v[i]-v1[j])+slove(i+1,j+1),slove(i,j+1));
    return d[i][j];

}
int main()
{
    long long int n,i,j,k,l,t,m,p;
    cin>>n;

    long long int f[n];
    //d[5001][5001]={-1};
    p=0;
    for(i=0;i<5001;i++)
    {
        for(j=0;j<5001;j++)
        {
            d[i][j]=-1;
        }
    }
    for(i=0;i<n;i++)
    {
        cin>>f[i];
    }
    for(i=0;i<n;i++)
    {
        if(f[i]==1)
        {
            v.push_back(i);
            p++;
        }
        else
        {
            v1.push_back(i);
        }
    }

    if(p==0)
    {
        cout<<0;
        return 0;
    }
    else
    {
        cout<<slove(0,0);

    }
}
