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;

constexpr int INF = (int) 1e9;

int main() {
    int n;
    cin >> n;
    vector<int> nums(n + 1);
    for (int i = 0; i < n; i++) {
        cin >> nums[i + 1];
    }

    vector<int> start_pos;
    for (int i = 0; i <= n; i++) {
        if (nums[i] == 1) {
            start_pos.push_back(i);
        }
    }

    int total_pep = (int) start_pos.size();
    if (total_pep == 0) {
        cout << 0 << endl;
        return 0;
    }
    vector<vector<int>> dp(n + 1, vector<int>(total_pep, INF));

    for (int i = 1; i <= n; i++) {
        if (nums[i] == 1) {
            dp[i][0] = dp[i - 1][0];
        }
        else {
            dp[i][0] = min(dp[i - 1][0], abs(start_pos[0] - i));
        }

    }

    for (int j = 1; j < total_pep; j++) {
        for (int i = 1; i <= n; i++) {
            if (i < j) {
                continue;
            }
            if (nums[i] == 1) {
                dp[i][j] = dp[i - 1][j];
            } else {
                dp[i][j] = min(dp[i - 1][j], dp[i - 1][j - 1] + abs(start_pos[j] - i));
            }
        }
    }
    // 1 0 0 1 0 0 1
    cout << dp[n][total_pep - 1] << endl;
}
