Kevin Sun wants to move his precious collection of n cowbells from Naperthrill to Exeter, where there is actually grass instead of corn. Before moving, he must pack his cowbells into k boxes of a fixed size. In order to keep his collection safe during transportation, he won't place more than two cowbells into a single box. Since Kevin wishes to minimize expenses, he is curious about the smallest size box he can use to pack his entire collection. 

Kevin is a meticulous cowbell collector and knows that the size of his i-th (1 ≤ i ≤ n) cowbell is an integer si. In fact, he keeps his cowbells sorted by size, so si - 1 ≤ si for any i > 1. Also an expert packer, Kevin can fit one or two cowbells into a box of size s if and only if the sum of their sizes does not exceed s. Given this information, help Kevin determine the smallest s for which it is possible to put all of his cowbells into k boxes of size s.

Input

The first line of the input contains two space-separated integers n and k (1 ≤ n ≤ 2·k ≤ 100 000), denoting the number of cowbells and the number of boxes, respectively.

The next line contains n space-separated integers s1, s2, ..., sn (1 ≤ s1 ≤ s2 ≤ ... ≤ sn ≤ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes si are given in non-decreasing order.

Output

Print a single integer, the smallest s for which it is possible for Kevin to put all of his cowbells into k boxes of size s.

Examples

Input

2 1
2 5


Output

7


Input

4 3
2 3 5 9


Output

9


Input

3 2
3 5 7


Output

8

Note

In the first sample, Kevin must pack his two cowbells into the same box. 

In the second sample, Kevin can pack together the following sets of cowbells: {2, 3}, {5} and {9}.

In the third sample, the optimal solution is {3, 5} and {7}.
import java.util.Scanner;

/**
 * Created by tignatchenko on 15.10.2015.
 */
public class Z1 {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();
        int[] s = new int[n];
        for (int i = 0; i<n; i++) {
            s[i] = sc.nextInt();
        }
        int k1 = k;
        int n1 = n;
        if (k1>=n) {
            k1 = n;
        } else
        while (k1!=n1 && k1>=0 && n1>=0) {
            n1-=2;
            k1-=1;
        }

        int c = k1;
        int m = 0;
        if (n==1) {
            m = s[0];
        } else if (c>=n) {
            m = s[n-1];
        } else if (c > 0) {
            if (c <= n) {
                int max = s[n-1];
                int n2 = n-c-1;
                for (int i = 0; i<(n-c)/2; i++) {
                    max = Math.max(s[i]+s[n2 - i], max);
                }
                m = max;
            } else {
                m = s[n-1];
            }
        } else {
            int max = s[0]+s[n-1];
            for (int i = 0; i<n/2; i++) {
                max = Math.max(s[i]+s[n-1 - i], max);
            }
            m = max;
        }

        System.out.println(m);

    }
}