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>
using namespace std;
#define intt long long int
#define F first 
#define S second
#define pb push_back

int main()
{
    int n;cin>>n;
    int arr[n+1];
    
    vector<int> pos;
    pos.pb(-1);

    for(int i=1;i<=n;i++) {cin>>arr[i];if(arr[i]==1) pos.pb(i); }
    int m=pos.size()-1; // no. of 1s

    int dp[n][m+1]; // dp[][]== ith posn tak use karke , pehle j 1's ko shift karna hai .

    for(int i=0;i<=n;i++)dp[i][0]=0;
    for(int i=1;i<=m;i++) dp[0][i]=1e9;

    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<=m;j++)
        {   
            if(arr[i]==1)
            dp[i][j]=dp[i-1][j];
            else 
            dp[i][j]=min(dp[i-1][j],dp[i-1][j-1]+abs(pos[j]-i));
        }
    }
    cout<<dp[n][m]<<endl;
    return 0;
}