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 pb push_back

typedef pair<int64_t, int64_t> pii;
typedef vector<int64_t> vi;
typedef vector<pii> vii;

void pre(){
    
}

bool multi = false;
int dp[2501][5001];
int mn[2501][5001];

void solve(){
    int n; cin>>n;
    memset(dp, 0, sizeof dp);
    memset(mn, 127, sizeof mn);
    vector<int> used, unused;
    for(int i=1; i <= n; i++){
        int x; cin>>x;
        if (x) used.push_back(i);
        else unused.push_back(i);
    }
    int nn=used.size(), mm=unused.size();
    int res = 0;
    for(int i=0; i <= mm; i++) mn[0][i] = 0;
    for(int i=0; i < nn; i++){
        for(int j = i; j < mm; j++){
            dp[i+1][j+1] = mn[i][j] + abs(used[i] - unused[j]);
        }
        for(int j = i; j < mm; j++){
            mn[i+1][j+1] = min(mn[i+1][j], dp[i+1][j+1]);
        }
    }
    // for(int i=0; i <= nn ;i++) {
    //     for(int j=0; j <= mm; j++){
    //         cout<<dp[i][j]<<" ";
    //     } cout<<endl;
    // }
    // cout<<endl;
    // for(int i=0; i <= nn ;i++) {
    //     for(int j=0; j <= mm; j++){
    //         cout<<mn[i][j]<<" ";
    //     } cout<<endl;
    // }
    cout<<mn[nn][mm];
}

int32_t main(){

#ifdef USE_FILE_IO
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

#ifndef USE_FILE_IO
    ios_base::sync_with_stdio(0); cin.tie();
#endif
    pre();
    int t = 1; 
    if (multi) cin>>t;
    while(t--) solve();
}