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;

void solve()
{

	int n;
	cin>>n;
	bool a[n]={0},occ[n]={0};
	vector<int> indz, indo;
	int i,j,ans=0;
	for(i=0; i<n; i++)
	{
		cin>>a[i];
		occ[i]=a[i];
		if(a[i]==1)
		{
			indo.push_back(i);
		}
		else
		{
			indz.push_back(i);
		}
	}

	int closedist,closeind,front,back,turns;	
	for(turns=1, front=indo.size()/2+1, back=indo.size()/2; turns<=indo.size(); turns++)
	{
		closedist=10000;
		if(turns&1==0&&back>=0 || front>=indo.size())
		{
			for(j=n-1; j>=0; j--)
			{
				if(occ[j]==0 && abs(indo[back]-j)<closedist)
				{
					closedist=abs(indo[back]-j);
					closeind=j;
				}
			}
			back--;
		}
		
		else
		{
			for(j=0; j<n; j++)
			{
				if(occ[j]==0 && abs(indo[front]-j)<closedist)
				{
					closedist=abs(indo[front]-j);
					closeind=j;
				}
			}
			front++;
		}
		occ[closeind]=1;
		ans+=closedist;
	}
	cout<<ans;
}


int main()
{

	#ifndef ONLINE_JUDGE
	freopen("inputcf.txt", "r", stdin);
	freopen("outputcf.txt", "w", stdout);
	#endif

	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	solve();

	return 0;

}