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 f1(a,b,c) for(int(a)=(b);(a)<(c);(a)++)
#define f2(a,b,c) for(int(a)=(b);(a)<=(c);(a)++)
#define f3(a,b,c) for(int(a)=(b);(a)>(c);(a)--)
#define f4(a,b,c) for(int(a)=(b);(a)>=(c);(a)--)
#define fi(a,b) for(auto(a)=b.begin();(a)!=b.end();(a)++)
#define pb push_back
#define all(v) v.begin(),v.end()
#define nper next_permutation
#define pper prev_permutation
#define mkp make_pair
#define F first
#define S second
#define int long long
#define mod 1000000007

typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<string> vs;
typedef vector<vs> vvs;
typedef vector<pii> vpii;
typedef set<int> si;
typedef unordered_set<int> usi;
typedef multiset<int> msi;
typedef map<int,int> mi;
typedef unordered_map<int,int> umi;
typedef multimap<int,int> mmi;

int32_t main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int n;
	cin>>n;
	vvi v(2);
	f1(i,0,n)
	{
		int x;
		cin>>x;
		v[x].pb(i);
	}
	int k=v[0].size();
	vvi dp(n-k+1,vi (k+1,0));
	f1(i,1,n-k+1)
	{
		f1(j,0,k+1)
		{
			if(j<i) dp[i][j]=INT_MAX;
			else dp[i][j]=min(dp[i][j-1],abs(v[1][i-1]-v[0][j-1])+dp[i-1][j-1]);
		}
	}
	cout<<dp[n-k][k];
	return 0;
}


















