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<algorithm>
#include<iostream>
#include<functional>
#include<cmath>
#include<map>
#include<cstdio>
#include<string.h>
#include<string>
#include<stack>
#include<vector>
#include<queue>
#include<set>
#include<iomanip>
#include<utility>
#define ll long long
#define ito ::iterator it
#define ioo ::itreator io
#define FOR(i,s,n) for(int i=s;i<n;++i)
using std::cin;
using std::cout;
using std::string;
using std::map;
using std::multiset;
using std::multimap;
using std::set;
//using std::min;
using std::stack;
using std::pair;
//using std::max;
using std::vector;
using std::queue;
using std::priority_queue;
typedef pair < int , int > ptt;
const int N = 1e5 + 5;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f;
const double eps = 1e-6;
int a[N],ca;
int b[N],cb;
int dp[5005][5005];
int main()
{
    //std::ios::sync_with_stdio(false);
    //std::cin.tie(0);std::cout.tie(0);
    int n; cin >> n;
    for(int i = 1; i <= n; i++){
        int x; cin >> x;
        if(x) b[++cb] = i;
        else a[++ca] = i;
    }
    for(int i = 1; i <= cb; i++) dp[0][i] = INF;
    for(int i = 1; i <= ca; i++){
        for(int j = 1; j <= cb; j++){
            if(i >= j){
                dp[i][j] = std::min(dp[i-1][j],dp[i-1][j-1] + abs(a[i] - b[j]));
            }
            else dp[i][j] = INF;
        }
    }
    cout << dp[ca][cb];
    //system("pause");
    return 0;
}