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;
const int p = 1e9 + 7;
const int lim = 1e6;
bool check(int elem) { return true; }
void bsearch() {
  int lo = 0, hi = 1e9, res = -1, mid;
  while (lo <= hi) {
    mid = lo + (hi - lo) / 2;
    if (check(mid)) {
      res = mid;
      hi = mid - 1;
    } else {
      lo = mid + 1;
    }
  }
}
struct Node {};
struct Compare {
  bool operator()(const Node &a, const Node &b) { return true; }
};
bool compare(const Node &a, const Node &b) { return true; }
void pre_io() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
int32_t main() {
  pre_io();
  int n;
  cin >> n;
  vector<int> v(n);
  for (int i = 0; i < n; i++) cin >> v[i];
  vector<int> t(v);
  long long res = 0;
  for (int i = n - 1; i >= 0; i--) {
    if (v[i] == 0) continue;
    int left = INT_MAX, right = INT_MAX;
    for (int j = i; j < n; j++) {
      if (t[j] == 0) {
        right = abs(i - j);
        break;
      }
    }
    for (int j = i; j >= 0; j--) {
      if (t[j] == 0) {
        left = abs(i - j);
        break;
      }
    }
    if (left < right) {
      t[i - left] = 1;
      res += left;
    } else {
      t[i + right] = 1;
      res += right;
    }
  }
  cout << res;
}
