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 <iostream>
#include <vector>

using namespace std;

const int inf = 1 << 30;

vector<int>freeChairs;
vector<int>occupiedChairs;
int n;

int dp[5007][5007];

int move(int current, int target) {
    return abs(occupiedChairs[current - 1] - freeChairs[target - 1]);
}

int main() {
    ios_base::sync_with_stdio(0);
    cin >> n;
    for(int i = 0; i < n; i++) {
        int x;
        cin >> x;
        if(x == 0) {
            freeChairs.push_back(i);
        } else {
            occupiedChairs.push_back(i);
        }
    }
    
    for(int i = 0; i <= n; i++) {
        for(int j = 0; j <= n; j++) {
            dp[i][j] = inf;
        }
    }
    
    dp[0][0] = 0;
    
    for(int i = 0; i <= occupiedChairs.size(); i++) {
        for(int j = 0; j <= freeChairs.size(); j++) {
            if(i > 0 && j > 0) {
                dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + move(i, j));
            }
            if(j > 0) {
                dp[i][j] = min(dp[i][j], dp[i][j - 1]);
            }
        }
    }
    
    int result = inf;
    for(int i = 0; i <= n; i++) {
        result = min(dp[occupiedChairs.size()][i], result);
    }
    
    cout << result << '\n';
    return 0;
}
