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 maxn = 5005;
namespace Standard_IO {
template <typename T>
void read(T &x) {
  T flag = 1;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') flag = -1;
  for (x = 0; isdigit(ch); ch = getchar()) x = x * 10 + ch - '0';
  x *= flag;
}
template <typename T>
void write(T x) {
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
template <typename T>
void print(T x) {
  if (x < 0) putchar('-'), x = -x;
  write(x);
  putchar(' ');
}
}  // namespace Standard_IO
using namespace Standard_IO;
int n;
int a[maxn], ans;
bool mark[maxn];
int stk1[maxn], stk2[maxn], tp1, tp2;
int calc() {
  tp1 = tp2 = 0;
  int ret = 0;
  for (int i = 1; i <= n; i++) {
    if (mark[i] == 1) stk1[++tp1] = i;
    if (a[i] == 1) stk2[++tp2] = i;
  }
  for (int i = 1; i <= tp1; i++) {
    ret += abs(stk1[i] - stk2[i]);
  }
  return ret;
}
void solve() {
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]), mark[i] = 0;
  for (int i = 1; i <= n; i++) {
    if (a[i] == 1) {
      for (int j = i - 1, k = i + 1;; j--, k++) {
        if (j > 0 && a[j] == 0 && !mark[j]) {
          mark[j] = 1;
          break;
        }
        if (k <= n && a[k] == 0 && !mark[k]) {
          mark[k] = 1;
          break;
        }
      }
    }
  }
  ans = calc();
  print(ans);
  puts("");
}
int main() {
  int T = 1;
  while (T--) solve();
  return 0;
}
