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 int long long 
int32_t main()
{
	ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	int n;
	cin>>n;
	int a[n];
	vector <int> v0;vector <int> v1;
	for(int i=0;i<n;++i) {cin>>a[i];if(!a[i]) v0.push_back(i);else v1.push_back(i);}
	int dp[v1.size()+1][v0.size()+1];
    for(int i=0;i<=v1.size();++i)
    {
    	for(int j=0;j<=v0.size();++j)
    	{
    		dp[i][j]=0;
    		if(i!=v1.size() && j==v0.size()) 
    		{
    			dp[i][j]=1e18;
    		}
    	}
    }
    for(int i=v1.size()-1;i>=0;i--)
    {
    	for(int j=v0.size()-1;j>=0;--j) 
    	{
    		dp[i][j]=min(dp[i][j+1],dp[i+1][j+1]+abs(v0[j]-v1[i]));
    	}
    }
    cout<<dp[0][0];
	return 0;
}