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 mod = 1000000007;
vector<vector<long long>> adj;
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) {
      res = res * a;
      res %= mod;
    }
    a = a * a;
    a %= mod;
    b >>= 1;
  }
  return res;
}
bool isPrime(int a) {
  if (a < 2) {
    return false;
  }
  if (a < 4) {
    return true;
  }
  if (a % 2 == 0 || a % 3 == 0) {
    return false;
  }
  for (long long i = 5; i * i <= a; i += 6) {
    if (a % i == 0 || a % (i + 2) == 0) {
      return false;
    }
  }
  return true;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  vector<long long> v1, v2;
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    if (v[i] == 0) {
      v2.push_back(i);
    } else {
      v1.push_back(i);
    }
  }
  if (v1.size() == 0) {
    cout << 0 << endl;
    return;
  }
  long long ans = 0;
  auto it = v2.end(), start = v2.begin();
  for (long long i = v1.size() - 1; i >= 0; i--) {
    auto it1 = upper_bound(start, it, v1[i]);
    if (it1 == it) {
      auto it2 = prev(it, 1);
      ans += abs(v1[i] - *it2);
      it = it2;
    } else {
      ans += abs(v1[i] - *it1);
      if (it1 != start) {
        it = it1;
      } else {
        start = next(it1, 1);
      }
    }
  }
  cout << ans << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
