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;
int n, a[100002], x[100002], y[100002], z;
bool vis = 1;
priority_queue<int> q;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1) {
      int j = 1, s = 0;
      while (!s) {
        if (i - j > 0)
          if (a[i - j] == 0)
            a[i - j] = -1, x[i] = i - j, y[i - j] = i, z += j, s = 1;
        if (i + j <= n && s == 0)
          if (a[i + j] == 0)
            a[i + j] = -1, x[i] = i + j, y[i + j] = i, z += j, s = 1;
        j++;
      }
    }
  }
  for (int u = 1; u <= n; u++)
    if (u < x[u]) q.push(u);
  while (!q.empty()) {
    int u = q.top();
    q.pop();
    int uu = x[u];
    if (u < uu) {
      for (int vv = 1; vv <= uu; vv++) {
        int v = y[vv], mx, mn, s = 0, f1 = 0, f2 = 0;
        if (v > vv && v > u) {
          vis = 1;
          x[u] = vv;
          y[vv] = u;
          x[v] = uu;
          y[uu] = v;
          mx = max(max(vv, uu), max(v, u));
          mn = min(min(vv, uu), min(v, u));
          if (u != mx && u != mn)
            if (!s)
              f1 = u, s = 1;
            else
              f2 = u;
          if (v != mx && v != mn)
            if (!s)
              f1 = v, s = 1;
            else
              f2 = v;
          if (uu != mx && uu != mn)
            if (!s)
              f1 = uu, s = 1;
            else
              f2 = uu;
          if (vv != mx && vv != mn)
            if (!s)
              f1 = vv, s = 1;
            else
              f2 = vv;
          if (f1 == 0 || f2 == 0) cout << -1;
          z -= 2 * abs(f1 - f2);
        }
      }
    }
  }
  cout << z;
}
