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;
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << '[';
  string sep;
  for (const auto &x : v) os << sep << x, sep = ", ";
  return os << ']';
}
template <typename T, size_t size>
ostream &operator<<(ostream &os, const array<T, size> &arr) {
  os << '{';
  string sep;
  for (const auto &x : arr) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '{' << p.first << ", " << p.second << '}';
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &s) {
  os << '{';
  string sep;
  for (const auto &x : s) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename A, typename B>
ostream &operator<<(ostream &os, const map<A, B> &mp) {
  os << '[';
  string sep;
  for (const auto &x : mp) os << sep << x, sep = ", ";
  return os << ']';
}
template <typename T>
ostream &operator<<(ostream &os, const multiset<T> &s) {
  os << '{';
  string sep;
  for (const auto &x : s) os << sep << x, sep = ", ";
  return os << '}';
}
template <typename T>
istream &operator>>(istream &in, vector<T> &v) {
  for (auto &i : v) in >> i;
  return in;
}
template <typename A, typename B>
istream &operator>>(istream &in, pair<A, B> &a) {
  in >> a.first >> a.second;
  return in;
}
void dbg_out() { cerr << endl; }
template <typename Head, typename... Tail>
void dbg_out(Head H, Tail... T) {
  cerr << ' ' << H;
  dbg_out(T...);
}
const string no = "NO";
const string yes = "YES";
const long long nax = 5000;
const long long nax2 = 2505;
vector<vector<long long>> dp(nax2, vector<long long>(nax, -1));
long long cnt = 0;
long long f(long long i, long long j, long long n, long long m,
            vector<long long> &ones, vector<long long> &zeroes) {
  if (i != n && j >= m) return 1e18;
  if (i >= n) {
    return 0;
  }
  if (dp[i][j] != -1) return dp[i][j];
  cnt++;
  long long ans = 1e18;
  long long it =
      lower_bound(zeroes.begin() + j, zeroes.end(), ones[i]) - zeroes.begin();
  for (long long ii = max({0LL, j, it - 40}); ii < min(it + 40, m); ii++) {
    ans = min(ans,
              abs(ones[i] - zeroes[ii]) + f(i + 1, ii + 1, n, m, ones, zeroes));
  }
  return dp[i][j] = ans;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> v(n);
  cin >> v;
  vector<long long> ones, zeroes;
  for (long long i = 0; i < n; i++) {
    if (v[i] == 1)
      ones.push_back(i);
    else
      zeroes.push_back(i);
  }
  sort(ones.begin(), ones.end()), sort(zeroes.begin(), zeroes.end());
  ;
  cout << f(0, 0, (long long)(ones).size(), (long long)(zeroes).size(), ones,
            zeroes)
       << '\n';
  ;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  for (long long tc = 1; tc <= t; tc++) {
    ;
    solve();
  }
  return 0;
}
