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;
using ll=long long;
#define                           pb                          push_back
#define                           fr(i,l,r)                   for(ll i=l;i<=r;i++)
#define                           rf(i,r,l)                   for(ll i=r;i>=l;i--)
#define                           done(i)                     cout<<"done = "<<i<<endl;
#define                           show(x,y)                   cout<<x<<" : ";for(auto z:y)cout<<z<<" ";cout<<endl;
#define                           fast                        ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define 						  all(x) 					  x.begin(),x.end()
#define 						  yes 	    				  cout<<"YES"<<endl
#define 						  no 	       				  cout<<"NO"<<endl
ll dp[5005][5005];
ll inf= 1e18;

void Test()
{

	int n;
	cin>>n;
	vector<ll>X,Y;
	X.pb(0);
	Y.pb(0);


	for(int i=1; i<=n; i++)
	{
		int x;
		cin>>x;
		if(x) X.pb(i);
		else Y.pb(i);
	}




	fr(i,1,n)
	fr(j,1,n)
	{
		dp[i][j]=inf;
		dp[i][0]=inf;

	}
	
	dp[0][0]=0;



	for(int i=1; i<X.size(); i++)
	{

		for(int j=1; j<Y.size(); j++)
		{

			ll cost= abs(X[i]-Y[j]) ;
			dp[i][j]= min(dp[i][j-1],dp[i-1][j-1]+ cost);    ;
		}
	}

	cout<<dp[X.size() - 1 ][Y.size() - 1]<<endl;









}


int main()
{

	int t=1;
	//cin>>t;

	while(t--)
		Test();

}


