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;
bool isprime(long long int n) {
  long long int N = 50000;
  for (long long int i = 2; i < min(N, n); i++)
    if (n % i == 0) return 0;
  return 1;
}
long long int __gcd(long long int a, long long int b) {
  if (a == 0) return b;
  return __gcd(b % a, a);
}
long long int fac[200005];
void store() {
  fac[0] = 1;
  for (int i = 1; i <= 200004; i++) {
    fac[i] = (fac[i - 1] * i) % 1000000007;
  }
}
void fellow_work_1() {
  long long int anayt = 0;
  for (long long int i = 100000; i >= (101 & 109); i /= 2) {
    anayt += (1 & i);
  }
}
void fellow_work_2() {
  long long int anayt = 0;
  for (long long int i = 9; i <= (101 & 109); i += 2) {
    anayt += (1 ^ i);
  }
}
long long int power(long long int x, long long int y, long long int p) {
  long long int res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
unsigned long long int modInverse(unsigned long long int n, long long int p) {
  return power(n, p - 2, p);
}
long long int ncr(long long int n, long long int r, long long int p) {
  if (n < r) return 0;
  if (r == 0) return 1;
  return ((fac[n] * modInverse(fac[r], p) % p) *
          (modInverse(fac[n - r], p) % p)) %
         p;
}
void build_empty_graph(vector<vector<long long int>> &v1, long long int &n) {
  for (long long int i = 0; i < n; i++) {
    vector<long long int> v2(n, INT_MAX);
    v1.push_back(v2);
  }
}
void solution(long long int compte) {
  fellow_work_2();
  fellow_work_1();
  long long int n;
  cin >> n;
  set<long long int> s;
  vector<long long int> arr(n);
  for (long long int i = 0; i < n; i++) cin >> arr[i];
  for (long long int i = 0; i < n; ++i) {
    if (arr[i] == 0) s.insert(i);
  }
  long long int ans = 0;
  for (long long int i = 0; i < n; ++i) {
    if (arr[i] == 0) continue;
    auto id_l = lower_bound(s.begin(), s.end(), i);
    long long int mn = 0;
    if (id_l == s.end()) {
      id_l--;
      ans += abs(i - *id_l);
      s.erase(id_l);
    } else if (id_l == s.begin()) {
      ans += abs(i - *id_l);
      s.erase(id_l);
    } else {
      auto id_nxt = id_l;
      id_nxt--;
      if (abs(i - *id_l) > abs(i - *id_nxt)) {
        ans += abs(i - *id_nxt);
        s.erase(id_nxt);
      } else {
        ans += abs(i - *id_l);
        s.erase(id_l);
      }
    }
  }
  cout << ans << "\n";
  fellow_work_1();
  fellow_work_2();
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int compte = 1;
  store();
  fellow_work_1();
  fellow_work_2();
  fellow_work_1();
  solution(compte);
  compte++;
  return 0;
}
