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 ll long long
ll dp[5001][5001];
vector<int> p, e;

ll solve(int i, int j){
    if(i >= p.size()){ // no person left, return 0
        return 0;
    }
    if(j >= e.size()){
        return 1e18;
    }
    if(dp[i][j] != -1){
        return dp[i][j];
    }
    // we have 2 options either to make ith person take jth chair or to make him take another chair
    // we will take the minimum of those 2 choices 
    dp[i][j] = min(solve(i, j+1), abs(e[j] - p[i]) + solve(i+1, j+1)); 
    return dp[i][j];
}

int main() {
	ios_base :: sync_with_stdio(false);
	cin.tie(NULL);
	int n;
	cin>>n;
	for(int i=0;i<n;i++){
	    int num;
	    cin>>num;
	    if(num){
	        p.push_back(i); // filled chairs
	    }
	    else{
	        e.push_back(i); // empty chairs
	    }
	}
	for(int i=0;i<n;i++){
	    for(int j=0;j<n;j++){
	        dp[i][j] = -1;
	    }
	}
	if(p.size() == 0){
	    cout<<"0";
	    return 0;
	}
	else{
	    cout<<solve(0, 0);
	}
	return 0;
}

