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.PriorityQueue;
import java.util.Scanner;


public class ParityAlternatedDeletions {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int[] a = new int[n];
        PriorityQueue<Integer> oddHeap = new PriorityQueue<>((x,y) -> y -x);
        PriorityQueue<Integer> evenHeap = new PriorityQueue<>((x,y) -> y -x);
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
            if ((a[i] & 1) == 0) evenHeap.add(a[i]);
            else oddHeap.add(a[i]);
        }

        int previousParity = -1;
        if(oddHeap.size() > evenHeap.size()) {
            oddHeap.poll();
            previousParity = 1;
        } else {
            evenHeap.poll();
            previousParity = 0;
        }

        while(!oddHeap.isEmpty() && !evenHeap.isEmpty()) {
            if (previousParity == 1) {
                evenHeap.poll();
                previousParity = 0;
            } else {
                oddHeap.poll();
                previousParity = 1;
            }
        }

        if (previousParity == 1 && !evenHeap.isEmpty()) {
            evenHeap.poll();
        } else if (previousParity == 0 && !oddHeap.isEmpty()) {
            oddHeap.poll();
        }

        if (oddHeap.isEmpty() && evenHeap.isEmpty()) {
            System.out.println(0);
        } else if(evenHeap.isEmpty()) {
            System.out.println(oddHeap.stream().reduce(0, (x, y) -> x + y));
        } else {
            System.out.println(evenHeap.stream().reduce(0, (x, y) -> x + y));
        }
    }
}