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>
#define ll long long
#define iPair pair<int, int>

using namespace std;

int n;
bool a[5001];
int dp[5001][5001]; /// dp[i][j] - choose from positions 1 to i exactly j finish positions

void mini(int &a, int b)
{
    a=min(a, b);
}

void solve()
{
    int i, j;
    vector<int> v;
    cin >> n;
    for(i=1; i<=n; i++)
    {
        cin >> a[i];
        if(a[i])
            v.push_back(i);
    }
    for(i=1; i<=n; i++)
        for(j=0; j<=i; j++)
            dp[i][j]=1e9;
    int sz=0;
    if(!v.empty())
        sz=v.size();
    for(i=0; i<n; i++)
        for(j=0; j<=i; j++)
        {
            /// don't choose position i+1
            if(j<=sz)
                mini(dp[i+1][j], dp[i][j]);

            /// choose position i+1
            if(!a[i+1] && j+1<=sz)
                mini(dp[i+1][j+1], dp[i][j]+abs(v[j]-(i+1)));
        }
    cout << dp[n][sz];
}

int main()
{
    ios::sync_with_stdio(0);
    cin.tie(0);
    int i, t;
    solve();
    return 0;
}
