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.Arrays;
import java.util.Scanner;

public class ParityAltDel {
	public static void main(String[] args) {

		Scanner in = new Scanner(System.in);
		int n = in.nextInt(), i, a[] = new int[n], e = 0, o = 0, t = 0, m = 0, x = 0;
		for (i = 0; i < n; i++) {

			a[i] = in.nextInt();
			if (a[i] % 2 == 0) {
				e++;
			} else {
				o++;
			}
		}
		in.close();
		Arrays.sort(a);

		if (e < o) {
			t = 1;
			m = o - e - 1;
		} else {
			m = e - o - 1;
		}
		for (i = 0; i < n && m > 0; i++)
			if (a[i] % 2 == t) {
				x += a[i];
				m--;
			}
		System.out.println(x);
	}
}
