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;
void init() {
  ios::sync_with_stdio(0);
  cin.tie(0);
}
void solve();
int main() {
  init();
  solve();
  return 0;
}
bool isprime(long long n) {
  for (long long i = 2; i < sqrt(n) + 1 && i < n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
void solve() {
  int n;
  cin >> n;
  vector<long long> odd;
  vector<long long> even;
  long long in;
  for (int i = 0; i < n; i++) {
    cin >> in;
    (in % 2 == 0) ? even.push_back(in) : odd.push_back(in);
  }
  if (!even.empty()) sort(even.begin(), even.end());
  if (!odd.empty()) sort(odd.begin(), odd.end());
  int countodd;
  int counteven;
  (!odd.empty()) ? countodd = odd.size() : countodd = 0;
  (!even.empty()) ? counteven = even.size() : counteven = 0;
  long long suma = 0;
  long long sumb = 0;
  if (countodd - counteven > 1) {
    for (int i = 0; i < countodd - counteven - 1; i++) {
      suma += odd[i];
    }
  }
  if (counteven - countodd > 1) {
    for (int i = 0; i < counteven - countodd - 1; i++) {
      sumb += even[i];
    }
  }
  cout << max(suma, sumb) << "\n";
}
