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;
//#include <ext/pb_ds/assoc_container.hpp>
//#include <ext/pb_ds/tree_policy.hpp>
//using namespace __gnu_pbds;
//template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag,tree_order_statistics_node_update>;
// order_of_key (k) : Number of items strictly smaller than k .
// find_by_order(k) : K-th element in a set (counting from zero).
const int mod =  1000000007;
#define int long long
#define FRE freopen("paint.in", "r", stdin),freopen("paint.out", "w", stdout);
#define IOS ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define all(a) a.begin(), a.end()
#define vi vector<int>
#define pb push_back
#define ss second
#define ff first

int power(int x, int n, int m=mod){
    int res = 1;
    while(n){
          if(n&1){
                res = res * x % m;
          }
          x = x * x % m;
          n>>=1;
    }
    return (res)%m;
}
/*ool prime[1000006];
void sieve(int n)
{
      memset(prime, true, sizeof(prime));
      int rootn = (int)sqrt(n);
      for (int p = 2; p <= rootn; p++)
            if(prime[p] == true)
                  for(int i = p*p; i <= n; i += p)
                        prime[i] = false;
      prime[1] = 0;

}*/

int dp[3000][5000];
int32_t main()
{
  IOS;
  int tt=1,n,k,x,y,z,i,j;
  //cin>>tt;
   for(int cas=1; cas<=tt; cas++){
     cin>>n;
     vi a;
     vi b;
     for(int i=0; i<n; i++){
       cin>>x;
       if(x==1)a.pb(i+1);
       else b.pb(i+1);
     }
     int s = a.size();
     if(s==0){
       cout<<0<<endl;
       continue;
     }
     int t = b.size();
     for(int i=0; i<=s; i++){
       for(int j=0; j<=t; j++){
         dp[i][j]=1e12;
       }
     }
    // dp[i][j]=min value for matching first i ppl using first j spaces and i matched to j
    vi ans(t+4);
     //for(int j=0; j<=t; j++)dp[0][j]=0;

     for(int i=1; i<=s; i++){

       for(int j=i; j<=t; j++){
         dp[i][j] = min(dp[i][j], abs(a[i-1]-b[j-1])+ans[j-1]);
       }

       ans[1]=dp[i][1];
       for(int j=2;j<=t; j++){
         ans[j]=min(ans[j-1],dp[i][j]);
       }
       //cout<<dp[i][t]<<endl;
     }
/*     for(int i=0; i<=s; i++){
       for(int j=0; j<=t; j++){
         cout<<dp[i][j]<<" ";
       }
       cout<<endl;
     }*/
     cout<<ans[t]<<endl;
    }
}
