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.
#pragma GCC optimize(2)
#pragma GCC optimize(3)
#include<bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define fio ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define mse(a,b) memset(a,b,sizeof a)
#define pb push_back
#define pii pair<int,int>
using namespace std;
const int mod=1e9+7;
const int maxx=1e6+10;
const double eps=1e-6;
using namespace std;
const ll inf=0x3f3f3f3f3f3f3f;
const long double PI = 3.14159265358979323846;
//inline ll read(){ ll x=0,f=1; char ch=getchar(); while (!isdigit(ch)) { if (ch=='-') f=-1; ch=getchar(); } while (isdigit(ch)) { x=x*10+ch-48;  ch=getchar();  } return x*f;}
// ll cc = ((1ll << 62) - 1 + (1ll << 62));
/*struct node {
    ll l,r ,lazy,val;
    node() :l(),r(),lazy(),val(){}
    node(ll a, ll b, ll c,ll d) :l(a),r(b),lazy(c),val(d){}

};*/
int a[maxx],b[maxx];
int dp[5050][5050];
signed main(){
            mse(dp,0x3f3f3f);
            dp[0][0]=0;
        //   mse(dp,0);
        int n; cin>>n;
        int a1,a0;a1=a0=0;
       for(int i=1;i<=n;i++)
       {dp[0][i]=0;
           int aa;cin>>aa;
           if(aa) a[++a1]=i;
           else b[++a0]=i;
       }
      for(int i=1;i<=a1;i++){
        for(int j=1;j<=a0;j++){
            dp[i][j]=min(dp[i-1][j-1]+abs(a[i]-b[j]),dp[i][j-1]);
        }
      }

    cout<<dp[a1][a0]<<endl;

  return 0;
}
