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.
// https://codeforces.com/blog/entry/90793?#comment-792051

#include <bits/stdc++.h>
#define ll long long
using namespace std;

ll fp(ll a, ll b){
    ll res=1;
    while(b){
        if(b&1)
        res*=a;
        a*=a;
        b/=2;
    }
    return res;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll n, m;
    cin>>n;
    vector<ll> a, b;
    for(int i=0;i<n;i++){
        ll x;
        cin>>x;
        if(x) a.push_back(i);
        else b.push_back(i);
    }
    n = a.size(); m = b.size();
    vector<vector<ll>> dp(n+1, vector<ll>(m+1, 1e9));
    for(ll i=0;i<=n;i++){
        for(int j=i;j<=m;j++){
            if(i == 0) dp[i][j] = 0;
            else if(j > 0){
                dp[i][j] = min(dp[i][j-1], dp[i-1][j-1] + abs(a[i-1] - b[j-1]));
            }
        }
    }
    cout<<dp[n][m];
	return 0;
}
