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 long long INF = (long long)0x3f3f3f3f3f3f3f, MAX = 9e18, MIN = -9e18;
const int N = 1e6 + 10, M = 2e6 + 10, mod = 1e9 + 7, inf = 0x3f3f3f3f;
long long a[N], c[N], n;
vector<long long> p1, pos;
bool vis[N], vis1[N];
void add(int i, int k) {
  while (i <= n) c[i] += k, i += i & -i;
}
long long get(int i) {
  int s = 0;
  while (i) s += c[i], i -= i & -i;
  return s;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int _;
  _ = 1;
  while (_--) {
    long long ans = 0;
    cin >> n;
    for (int i = 1; (i) <= (n); i++) {
      cin >> a[i];
      if (a[i])
        p1.push_back(i), vis1[i] = 1;
      else
        add(i, 1);
    }
    int n1 = p1.size();
    for (int i = 0; (i) < (n1); i++) {
      int j1 = p1[i] - 1, j2 = p1[i] + 1;
      while (j1 >= 1 && (vis[j1] || vis1[j1])) j1--;
      while (j2 <= n && (vis[j2] || vis1[j2])) j2++;
      if (j1 == 0) {
        pos.push_back(j2);
        vis[j2] = 1;
        add(j2, -1);
        continue;
      }
      if (j2 == n + 1) {
        pos.push_back(j1);
        vis[j1] = 1;
        add(j1, -1);
        continue;
      }
      if (abs(p1[i] - j1) <= abs(p1[i] - j2) || get(n) - get(j2 - 1) < n1 - i) {
        pos.push_back(j1);
        vis[j1] = 1;
        add(j1, -1);
      } else {
        pos.push_back(j2);
        vis[j2] = 1;
        add(j2, -1);
      }
    }
    sort(pos.begin(), pos.end());
    for (int i = 0; (i) < (n1); i++) {
      ans += abs(p1[i] - pos[i]);
    }
    cout << ans << '\n';
  }
  return 0;
}
