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
import java.util.*;

public class Main {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int n = s.nextInt();
        int[] a = new int[n];
        List<Integer> t1 = new ArrayList<>();//偶数列
        List<Integer> t2 = new ArrayList<>();//奇数列
        for(int i=0;i<n;i++){
            a[i]=s.nextInt();
        }
        for(int i=0;i<a.length;i++){
            if(a[i]%2==0){
                t1.add(a[i]);
            }else {t2.add(a[i]);}
        }
        //倒叙排序
        Collections.sort(t1);
        Collections.sort(t2);
        Collections.reverse(t1);
        Collections.reverse(t2);

        if(t1.size()>=t2.size()){
            if(t1.size()-t2.size()<=1){
                System.out.println(0);
            }else {
                int sum =0 ;
                for(int i=t2.size()+1;i<t1.size();i++){
                    sum+=t1.get(i);
                }
                System.out.println(sum);
            }
        }
            if(t2.size()>t1.size()){
                if(t2.size()-t1.size()==1){
                    System.out.println(0);
                }else {
                    int sum =0 ;
                    for(int i=t1.size()+1;i<t2.size();i++){
                        sum+=t2.get(i);
                    }
                    System.out.println(sum);
                }
        }
    }
}