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 GO ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define int long long int
#define endl "\n"
const int maxr = 5e3 + 5;
const int INF = 1e12;
int arr[maxr], n;
vector<int> ones;
int dp[maxr][maxr];

int solve(int idx, int oind) {

    if(oind == (int)ones.size()) {
        return 0;
    }
    if(idx == n) {
        return INF;
    }
    int &ans = dp[idx][oind];
    if(ans != -1)
        return ans;

    if(arr[idx] == 0) {
        int a = solve(idx + 1, oind + 1) + abs(ones[oind] - idx);
        int b = solve(idx + 1, oind);
        return ans = min(a, b);
    }
    return ans = solve(idx + 1, oind);
}

int32_t main()
{
    GO;
    cin >> n;
    for(int i = 0; i < n; i++) {
        cin >> arr[i];
        if(arr[i])
            ones.push_back(i);
    }

    memset(dp, -1, sizeof(dp));
    int answer = solve(0, 0);
    cout << answer << endl;
    return 0;
}