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 haiyaku ios::sync_with_stdio(false), cin.tie(NULL);
#define PB push_back
#define F first
#define S second
#define mp make_pair
#define all(x) x.begin(), x.end()
#define int long long
//#define double long double
 
const int MINI = -2147483647;
const int MAXI = 2147483647;
const long long MAXLL = 9223372036854775807ll / 4;
const long long MINLL = -9223372036854775807ll / 4;
 
using namespace std;
 
using pii = pair<int, int>;
using pll = pair<long long, long long>;


signed main () {
  ios::sync_with_stdio(false), cin.tie(NULL);

  int n; cin >> n;
  vector<int> a(n);
  for(int &X : a) cin >> X;

  vector<int> pos(0);
  for(int i = 0; i < n; i++) if (a[i] == 1) pos.PB(i);

  vector<int> dp(pos.size() + 1, MAXLL);
  dp[0] = 0;
  for(int i = 0; i < n; i++) {
    if (a[i] == 0) {
      for(int j = pos.size(); j > 0; j--) {
        dp[j] = min(dp[j], dp[j - 1] + abs(pos[j - 1] - i));
      }
    } 
  } 

  cout << dp.back() << '\n';

  return 0;
}
