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;
vector<bool> prime;
vector<long long> fact, inv, primes, factors;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
void factorize(long long a) {
  factors.clear();
  for (long long i = 1; i * i <= a; i++) {
    if (i * i == a)
      factors.push_back(i);
    else if (a % i == 0) {
      factors.push_back(i);
      factors.push_back(a / i);
    }
  }
  sort(factors.begin(), factors.end());
}
long long power(long long a, long long b) {
  if (a == 1 || b == 0) return 1;
  long long c = power(a, b / 2);
  long long res = 1;
  res = c * c;
  if (res >= 1000000007) res %= 1000000007;
  if (b % 2) res *= a;
  if (res >= 1000000007) res %= 1000000007;
  return res;
}
long long modInv(int a) { return power(a, 1000000007 - 2); }
void factorial(long long n) {
  fact.resize(n + 1);
  fact[0] = 1;
  for (long long i = 1; i <= n; i++) {
    fact[i] = fact[i - 1] * i;
    if (fact[i] >= 1000000007) fact[i] %= 1000000007;
  }
}
void InvFactorial(long long n) {
  inv.resize(n + 1);
  inv[0] = 1;
  for (long long i = 1; i <= n; i++) inv[i] = modInv(fact[i]);
}
long long ncr(long long n, long long r) {
  if (n < r || n < 0 || r < 0) return 0;
  long long b = inv[n - r];
  long long c = inv[r];
  long long a = fact[n] * b;
  if (a >= 1000000007) a %= 1000000007;
  a *= c;
  if (a >= 1000000007) a %= 1000000007;
  return a;
}
void remove_duplicates(vector<pair<long long, long long> > &v) {
  sort(v.begin(), v.end());
  long long _size = unique(v.begin(), v.end()) - v.begin();
  v.resize(_size);
}
unsigned long long gcd(unsigned long long u, unsigned long long v) {
  if (u == 0 || v == 0) return max(u, v);
  unsigned long long shift = __builtin_ctz(u | v);
  u >>= __builtin_ctz(u);
  do {
    v >>= __builtin_ctz(v);
    if (u > v) swap(u, v);
    v -= u;
  } while (v != 0);
  return u << shift;
}
void sieve(int n) {
  prime.assign(n + 1, 1);
  prime[1] = false;
  for (long long p = 2; p * p <= n; p++)
    if (prime[p])
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
  for (long long i = 2; i < n + 1; i++)
    if (prime[i]) primes.push_back(i);
}
long long size(long long l) {
  long long p = l;
  long long count = 0;
  while (p) p /= 10, count++;
  return count;
}
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  set<long long> s, t;
  for (long long i = 0; i < n; i++) {
    if (a[i])
      s.insert(i + 1);
    else
      t.insert(i + 1);
  }
  long long ans = 0;
  while (!s.empty()) {
    auto i1 = s.end();
    i1--;
    auto i2 = t.upper_bound(*i1);
    if (i2 == t.end()) i2--;
    auto i3 = t.lower_bound(*i1);
    if (i3 == t.end()) i3--;
    if (abs(*i2 - *i1) >= abs(*i3 - *i1)) {
      ans += abs(*i1 - *i3);
      s.erase(i1);
      t.erase(i3);
    } else {
      ans += abs(*i1 - *i2);
      s.erase(i1);
      t.erase(i2);
    }
  }
  cout << ans << '\n';
}
signed main() {
  fast();
  long long T = 1;
  for (long long i = 1; i <= T; i++) solve();
}
