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 PolyCrap {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
	    long a[] = new long[n];
		int e = 0,o=0;
		for(int i=0;i<n;i++) {
			a[i] = sc.nextLong();
			if(a[i]%2==0)
				e++;
			else
				o++;
		}
		Arrays.sort(a);
		long even[] = new long[e];
		long odd[] = new long[o];
		int j=0,k=0;
		for(int i=0;i<n;i++) {
			if(a[i]%2==0)
				even[j++]=a[i];
			else
				odd[k++]=a[i];
		}
		Arrays.sort(even);
		Arrays.sort(odd);
		int ev=even.length;
		int od=odd.length;
		
		if(ev==od || ev==od+1 || od==ev+1)
			System.out.println(0);
		else if(ev>od+1) {
			long sum=0;
			for(int m=0;m<(ev-od-1);m++) {
				sum+=even[m];
			}
			System.out.println(sum);
		}
		else if(od>ev+1) {
			long sum=0;
			for(int m=0;m<(od-ev-1);m++) {
				sum+=odd[m];
			}
			System.out.println(sum);
		}
	}

}
