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.
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#include<bits/stdc++.h>

#define all(x) x.begin(),x.end()
#define pb push_back
#define mp make_pair
#define Unique(x) x.erase(unique(all(x)),x.end());

using namespace std;

const int dx[]={-1,0,1,0};
const int dy[]={0,-1,0,1};

const int Days[]={-1,31,28,31,30,31,30,31,31,30,31,30,31};

inline int read()
{
	#define gc getchar
	int ans=0;
	bool f=1;
	char ch=gc();
	
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	
	while(isdigit(ch))ans=((ans<<2)+ans<<1)+(ch^48),ch=gc();
	
	return f?ans:-ans;
	
	#undef gc
}

typedef long long ll;

vector<int> a,b;
int n;
int dp[5010];

int main(int argc, char const *argv[])
{
	ios::sync_with_stdio(false);
	
	a.pb(-1);
	b.pb(-1);
	
	cin>>n;for(int i=1;i<=n;i++)
	{
		int x;cin>>x;
		
		if(x)a.pb(i);
		else b.pb(i);
	}
	
	memset(dp,0x3f,sizeof(dp));
	dp[0]=0;
	
	int cnta=a.size();
	int cntb=b.size();
	
	for(int i=1;i<cntb;i++)for(int j=min(cnta-1,i);j>=1;j--)dp[j]=min(dp[j],dp[j-1]+abs(a[j]-b[i]));
	
	cout<<dp[cnta-1];	
	
	//printf("Time used = %.12f",(double)(clock())/CLOCKS_PER_SEC);
	return 0;
}

/*  GOD BLESS  */
