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;

int get_min(const vector<pair<int, int>>& v, const int& x) {
    for(const auto& pp : v) {
        if (pp.second <= x) { return pp.first; }
    }
}

int main()
{
    ios_base::sync_with_stdio(false); cin.tie(0);
    int n, val;
    cin >> n;
    vector<int> z; z.reserve(n);
    vector<int> u; u.reserve(n);
    for (int i=0; i<n; i++) {
        cin >> val;
        if (val == 1)
            u.push_back(i);
        else
            z.push_back(i);
    }

    const int zsz = static_cast<int>(z.size());
    const int usz = static_cast<int>(u.size());

    const int dpsz = zsz - usz + 1;
    vector<int> dp(dpsz, 0);

    vector<pair<int, int>> sorted_dp(dpsz);

    for(int i=0; i<usz; i++) {
        for(int k=0; k<dpsz; k++) {
            sorted_dp[k] = {dp[k], k};
        }
        sort(sorted_dp.begin(), sorted_dp.end());

        for(int j=i+dpsz-1; j>=i; j--) {
            dp[j-i] = get_min(sorted_dp, j-i);
            dp[j-i] += abs(u[i] - z[j]);
        }
    }

    val = *(min_element(dp.begin(), dp.end()));
    cout << val;
    return 0;
}
