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 int long long
#define rapido ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define endl "\n"
#define MAXN 5005
#define INF 1e9


int dp[MAXN][MAXN];

void solve()
{
   int n;cin>>n;
   vector<int> black;vector<int> white;
        for(int i=0;i<n;i++) 
        {
            int a;cin >> a;
            if(a == 0) 
                black.push_back(i);
            else
                white.push_back(i);
        }
        for(int i=0;i<MAXN;i++)
        {
            for(int j=0;j<MAXN;j++)
            dp[i][j]=INF;
        }
        dp[0][0] = 0;
        for(int i=0;i<=white.size();i++)
        {
            for(int j=0;j<=black.size();j++) 
            {
                if(i == 0) 
                {
                    dp[i][j] = 0;
                    continue;
                }
                if(j == 0) 
                    continue;
                dp[i][j] = min(dp[i][j-1], dp[i-1][j-1] + abs(white[i-1] - black[j-1])); 
            }
        }
        cout << dp[white.size()][black.size()] << endl;
}
int32_t main() {
    rapido;
    int t=1;
    //cin>>t;
    while(t--) 
    solve();
	return 0;
}
