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 ll long long
#define IO ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
using namespace std;
double PI=acos(-1);

int n,m;
ll dp[5005][5005];
int pos[5005],good[5005];
ll solve(int idx,int last)
{
    if(idx==n) return 0;
    if(last==m) return 1e17;
    if(dp[idx][last]!=-1)
        return dp[idx][last];
    ll ret=1e10;
    ret=min(ret,solve(idx+1,last+1)+abs(good[last]-pos[idx]));
    ret=min(ret,solve(idx,last+1));
    return dp[idx][last]=ret;
}

void test_case()
{
    cin>>n;
    int idx=0,last=0;
    memset(dp,-1,sizeof dp);
    for(int i=0;i<n;i++)
    {
        int x;cin>>x;
        if(x==1)
        {
            pos[idx]=i;
            idx++;
        }
        else
        {
            good[last]=i;
            last++;

        }
    }
    n=idx,m=last;
    cout<<solve(0,0)<<'\n';




}


int main(){
    IO
    int t;t=1;
    while(t--)
    {
        test_case();
    }





    return 0;
}
