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 FAST ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define int long long 
#define pi 3.14159265359
#define all(v) v.begin(),v.end()
#define vil vector<long long int>
#define vi vector<int> 
#define vb vector<bool>
#define pb(x) push_back(x)
#define pf(x) push_front(x)
#define mxe(v) *max_element(v.begin(),v.end())
#define mne(v) *min_element(v.begin(),v.end())
#define ub upper_bound
#define lb lower_bound
#define mod 1000000007
#define mxn 1000000000000000000
#define sec second
#define fir first
#define pil pair<long,long>
#define acc(v) accumulate(v.begin(),v.end(),0)
#define vi2 vector<vector<int>>
#define re
#define fir first
#define sec second
int find_set(int v,vi &p){
    if(v==p[v])
        return v;
    return p[v]=find_set(p[v],p);
}
void union_set(int a,int b,vi &p){
    a=find_set(a,p);
    b=find_set(b,p);
    if(a!=b){
        p[b]=a;
    }
}
bool mysort(pair<int,int> a,pair<int,int> b){
    return a.fir<b.fir;
} 
void solve(){
int n;
cin>>n;
vi a(n),b(n,1);
for(int i=0;i<n;i++){
    cin>>a[i];
}
int ans=0;
for(int i=0;i<n;i++){
    if(a[i]==1){
        int p=4*n;
        int q=4*n;
        for(int j=i-1;j>=0;j--){
            if(a[j]==0&&b[j]==1){
               p=j;
               break;
            }
        }
          for(int j=i+1;j<n;j++){
            if(a[j]==0&&b[j]==1){
               q=j;
               break;
            }
        }
        if(abs(p-i)<=abs(q-i)){
          b[p]=0;
           ans+=abs(p-i);
        }
        else if(abs(p-i)>abs(q-i)){
            b[q]=0;
            ans+=abs(q-i);
        }
    }
}
cout<<ans<<endl;
return;






}
signed main() 
{ 
    FAST;
    int t;
    t=1;
   //cin>>t;
    while(t--){
        solve();
    }
    return 0; 
} 