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>
using namespace std;

template<typename T, typename U> static inline void amin(T &x, U y){ if(y<x) x=y; }
template<typename T, typename U> static inline void amax(T &x, U y){ if(x<y) x=y; }

void domain_expansion_() {
  int n;
  cin >> n;
  vector<int> emp, f;
  for (int i = 0, a; i < n; i++) {
    cin >> a;
    if (!a) emp.push_back(i);
    else f.push_back(i);
  }
  int F = (int) f.size(), P = (int) emp.size();
  vector<vector<int>> dp(F + 1, vector<int> (P + 1, 1e9));
  for (int i = 0; i <= F; i++) {
    for (int j = 0; j <= P; j++) {
      if (i == 0) {
        dp[i][j] = 0;
        continue;
      }
      if (j == 0) continue;
      amin(dp[i][j], dp[i - 1][j - 1] + abs(f[i - 1] - emp[j - 1]));
      amin(dp[i][j], dp[i][j - 1]);
    }
  }
  cout << dp[F][P];
  return;
}  
  
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tt = 1;
  // cin >> tt;
  while (tt--) 
    domain_expansion_();
  return 0;
}