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 inf 1001000009
#define infmx 1e18
#define ff first
#define ss second
#define ll long long
#define pb push_back
#define MP make_pair
#define IOS ios_base::sync_with_stdio(0);cin.tie(NULL);cout.tie(NULL)
#define all(x) x.begin(),x.end()

typedef pair<int,int>pi;
typedef pair<long long,long long>pll;
typedef pair<pi,int>ppi;
typedef pair<int,pi>pip;
typedef vector<int>vi;
typedef vector<pi>vpi;
const int mod=1e9+7;
const int N=5005;
int ara[N];
int dp[N][N];
int one[N]; ///position of ith one
int k,n;
int call(int i,int tk)
{
    if(tk==k)return 0;
    if(i>=n+1)return inf;
    if(dp[i][tk]!=-1)return dp[i][tk];
    int f1 = inf,f2=inf;
    if(ara[i]==0)       ///take the ith 0
        f1 = abs(i-one[tk+1])+call(i+1,tk+1);
    f2 = call(i+1,tk);
    return dp[i][tk]=min(f1,f2);
}
int main()
{
    IOS;
    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>ara[i];
        if(ara[i]==1)
            one[++k] = i;
    }
    memset(dp,-1,sizeof dp);
    int ans = call(1,0);
    cout<<ans<<endl;
}

