Polycarp has an array a consisting of n integers.

He wants to play a game with this array. The game consists of several moves. On the first move he chooses any element and deletes it (after the first move the array contains n-1 elements). For each of the next moves he chooses any element with the only restriction: its parity should differ from the parity of the element deleted on the previous move. In other words, he alternates parities (even-odd-even-odd-... or odd-even-odd-even-...) of the removed elements. Polycarp stops if he can't make a move.

Formally: 

  * If it is the first move, he chooses any element and deletes it; 
  * If it is the second or any next move: 
    * if the last deleted element was odd, Polycarp chooses any even element and deletes it; 
    * if the last deleted element was even, Polycarp chooses any odd element and deletes it. 
  * If after some move Polycarp cannot make a move, the game ends. 



Polycarp's goal is to minimize the sum of non-deleted elements of the array after end of the game. If Polycarp can delete the whole array, then the sum of non-deleted elements is zero.

Help Polycarp find this value.

Input

The first line of the input contains one integer n (1 ≤ n ≤ 2000) — the number of elements of a.

The second line of the input contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 10^6), where a_i is the i-th element of a.

Output

Print one integer — the minimum possible sum of non-deleted elements of the array after end of the game.

Examples

Input


5
1 5 7 8 2


Output


0


Input


6
5 1 2 4 6 3


Output


0


Input


2
1000000 1000000


Output


1000000
#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  long long p = 1;
  while (b) {
    if (b % 2) {
      p = p * a;
    }
    b >>= 1;
    a = a * a;
  }
  return p;
}
bool prime(long long n) {
  if (n <= 3) {
    return true;
  }
  for (long long i = 2; i <= sqrt(n); ++i) {
    if (n % i == 0) {
      return true;
    }
  }
  return false;
}
long long gcd(long long a, long long b) {
  if (a == 0) {
    return b;
  }
  return gcd(b % a, a);
}
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  vector<long long> even, odd;
  long long ans = 0;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    ans += a[i];
    if (a[i] % 2) {
      odd.push_back(a[i]);
    } else
      even.push_back(a[i]);
  }
  if (odd.size() == 0) {
    sort(even.begin(), even.end());
    cout << ans - even[even.size() - 1] << "\n";
    return;
  }
  if (even.size() == 0) {
    sort(odd.begin(), odd.end());
    cout << ans - odd[odd.size() - 1] << "\n";
    return;
  }
  sort(odd.begin(), odd.end());
  sort(even.begin(), even.end());
  long long i = odd.size() - 1, j = even.size() - 1;
  if (abs(i - j) <= 1) {
    cout << 0 << "\n";
    return;
  }
  if (i > j) {
    ans -= odd[i];
    i--;
  } else {
    ans -= even[j];
    j--;
  }
  long long x = min(i, j) + 1;
  long long y = x;
  while (y > 0) {
    ans -= odd[i];
    i--;
    y--;
  }
  while (x > 0) {
    ans -= even[j];
    j--;
    x--;
  }
  cout << ans << "\n";
  return;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}
