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 rahul ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define ll long long
#define ld long double
#define pb push_back
#define mp make_pair
#define sp setprecision
#define mn INT_MIN
#define mx INT_MAX
#define F first
#define S second
#define gcd __gcd
#define all(a) a.begin(), a.end()
#define FOR(i,a,b) for(int i=(a);i<=(b);++i)
#define ROF(i,a,b) for(int i=(a);i>=(b);--i)
const long mod=pow(10,9)+7;
const double pi=3.141592653589793238;


vector<ll>v,v1;
ll dp[5001][5001];

ll subprob(ll i,ll j)
{
    if(i>=v.size())return 0;
    if(j>=v1.size())return INT_MAX;
    if(dp[i][j]!=-1)
    {
        return dp[i][j];
    }
    dp[i][j] = min(abs(v[i]-v1[j])+subprob(i+1,j+1),subprob(i,j+1));
    return dp[i][j];
}

int main()
{
    rahul
    /*ll t;
    cin>>t;
    while(t--)
    {*/
        ll n,i,j=0;
        cin>>n;
        memset(dp,-1,sizeof(dp));
        ll a[n];
        for(i=0;i<n;i++)
        {
            cin>>a[i];
            if(a[i])
            {
                j++;
                v.pb(i);
            }
            else
            {
                v1.pb(i);
            }
        }
        if(j==0)
        {
            cout<<0<<endl;
        }
        else{
            cout<<subprob(0,0)<<endl;
        }
    //}
}
