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 <map>
#include <set>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define ll long long
#define ull unsigned long long
#define cint const int&
#define Pi acos(-1)

const int mod = 1e9+7;
const int inf_int = 0x3f3f3f3f;
const ll inf_ll = 0x7fffffffffffffff;
const double ept = 1e-9;

int n;
int a[5050], ct1;
int b[5050], ct2;
int dp[5050][5050];

int main() {
    cin >> n;
    int st;
    for(int i=1; i<=n; i++) {
        cin >> st;
        if(st) a[++ct1] = i;
        else b[++ct2] = i;
    }
    memset(dp, 0x3f, sizeof dp);
    memset(dp[0], 0, sizeof dp[0]);
    int ans = inf_int;
    for(int i=1; i<=ct1; i++)
        for(int j=1; j<=ct2; j++) {
            if(dp[i-1][j-1] < inf_int) dp[i][j] = min(dp[i][j], dp[i-1][j-1] + abs(a[i]-b[j]));
            dp[i][j] = min(dp[i][j], dp[i][j-1]);
            if(i == ct1) ans = min(ans, dp[i][j]);
        }
    // for(int i=1; i<=ct1; i++) {
    //     for(int j=1; j<=ct2; j++)
    //         cout << dp[i][j] << ' ';
    //     cout << endl;
    // }
    if(ans == inf_int) ans = 0;
    cout << ans << endl;

    return 0;
}