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.
#pragma GCC optimize("O2")
#include <bits/stdc++.h>
using namespace std;

#define ll     long long
#define test   int tt; cin>>tt; while(tt--)
#define ff     first
#define ss     second
#define pb     push_back
#define ppb    pop_back
#define pf     push_front
#define ppf    pop_front

int dp[5001][5001];

ll int cal(vector<int> &zero_pos, vector<int> &one_pos, int i, int j)
{
    if(i==zero_pos.size() && j<one_pos.size())
        return INT_MAX;

    if(j==one_pos.size())
        return 0;

    if(dp[i][j]!=-1)
        return dp[i][j];

    ll int x=fabs(zero_pos[i]-one_pos[j]);

    return dp[i][j]=min(cal(zero_pos, one_pos, i+1, j),
                        x+cal(zero_pos, one_pos, i+1, j+1));
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);

    fill(dp[0], dp[0]+5001*5001, -1);

    int n;
    cin>>n;

    int arr[n];
    for(int i=0; i<n; i++)
        cin>>arr[i];

    vector<int> zero_pos, one_pos;

    for(int i=0; i<n; i++)
    {
        if(arr[i]==0)
            zero_pos.pb(i);
        else
            one_pos.pb(i);
    }

    cout<<cal(zero_pos, one_pos, 0, 0)<<endl;
}
