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;
long long int min(long long int a, long long int b) { return a < b ? a : b; }
long long int min(long long int a, long long int b, long long int c) {
  a = min(a, b);
  a = min(a, c);
  return a;
}
long long int max(long long int a, long long int b) { return a > b ? a : b; }
long long int max(long long int a, long long int b, long long int c) {
  a = max(a, b);
  a = max(a, c);
  return a;
}
long long int dsum(long long int n) {
  long long int ans = 0;
  while (n) {
    ans += (n % 10);
    n /= 10;
  }
  return ans;
}
long long int power(long long int a, long long int b) {
  long long int res = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
long long int power(long long int a, long long int b, long long int m) {
  if (b == 0) return 1;
  if (b == 1) return a % m;
  long long int res = 1;
  a = a % m;
  while (b) {
    if (b & 1) {
      res = (res * a) % m;
    }
    a = ((a % m) * (a % m)) % m;
    b >>= 1;
  }
  return res;
}
long long int Ceil(long long int a, long long int b) {
  long long int res = a / b;
  if (a % b != 0) res++;
  return res;
}
template <typename T>
istream &operator>>(istream &is, vector<T> &v) {
  for (T &t : v) is >> t;
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  for (const T &t : v) {
    os << t << " ";
  }
  os << '\n';
  return os;
}
long long int dxx[] = {1, 1, 0, -1, -1, -1, 0, 1};
long long int dyy[] = {0, 1, 1, 1, 0, -1, -1, -1};
long long int dx[] = {1, 0, -1, 0};
long long int dy[] = {0, 1, 0, -1};
struct pll {
  long long int x, y;
};
bool comp(pll a, pll b) {
  if (a.x == b.x)
    return a.y < b.y;
  else
    return a.x < b.x;
}
void solve(long long int the) {
  char ch;
  double aa, bb, cc, dd, xx;
  string s1, s2, s3, s4, str;
  long long int i, j, k, a, b, c, d, n, m, l, r, x, y, z, low, high, mid,
      sum = 0, ans = 0, temp = 0, t, q;
  cin >> n;
  vector<long long int> v(n);
  cin >> v;
  set<long long int> se;
  for (long long int i = 0; i < n; i++)
    if (v[i] == 0) se.insert(i);
  for (long long int i = 0; i < n; i++) {
    if (v[i] == 0) continue;
    long long int id = i;
    auto it = se.lower_bound(id);
    if (it == se.begin()) {
      x = *it;
      ans += abs(id - x);
      se.erase(it);
    } else if (it == se.end()) {
      it--;
      x = *it;
      ans += abs(id - x);
      se.erase(it);
    } else {
      auto it1 = it;
      it1--;
      long long int diff1, diff2;
      x = *it;
      y = *it1;
      diff1 = abs(id - x);
      diff2 = abs(id - y);
      if (diff1 < diff2) {
        ans += diff1;
        se.erase(it);
      } else {
        ans += diff2;
        se.erase(it1);
      }
    }
  }
  cout << ans;
}
int32_t main() {
  long long int t = 1, c = 1;
  while (t--) {
    solve(c);
    c++;
  }
  return 0;
}
