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>
typedef long long ll;

#define pb push_back
#define _io ios_base::sync_with_stdio(false); cin.tie(0); cout.tie(0);
using namespace std;

#define PROB "Armchairs"
#define MAXN 5005

int a[MAXN];
ll dp[MAXN][MAXN];

int main(){
    _io
    //freopen(PROB".inp", "r", stdin);
    //freopen(PROB".out", "w", stdout);
    int n; cin >> n;
    assert(2 <= n && n <= 5000);
    vector <int> arr[2];
    arr[0].pb(0);
    arr[1].pb(0);
    for(int i = 1; i <= n; i++){
        cin >> a[i];
        assert(0 <= a[i] && a[i] <= 1);
        arr[a[i]].pb(i);
    }
    int m1 = arr[1].size();
    int m2 = arr[0].size();
    for(int i = 0; i <= n; i++)
        for(int j = 0; j <= n; j++)
            dp[i][j] = INT_MAX;
    for(int i = 0; i < m2; i++)
        dp[0][i] = 0;
    for(int i = 1; i < m1; i++)
        for(int j = 1; j < m2; j++){
            int u = arr[1][i], v = arr[0][j];
            dp[i][j] = dp[i][j-1];
            dp[i][j] = min(dp[i][j], dp[i-1][j-1] + abs(u-v));
        }
    cout << dp[m1-1][m2-1];
}
