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;

public class Solution {

    public static void main(String[] args) {

        Scanner s = new Scanner(System.in);

        String[] cowbellsAndBoxes = s.nextLine().split(" ");
        int cb = Integer.valueOf(cowbellsAndBoxes[0]);
        int boxes = Integer.valueOf(cowbellsAndBoxes[1]);

        String[] bs = s.nextLine().split(" ");
        int[] bells = new int[cb];
        for (int i = 0; i < bells.length; i++) {
            bells[i] = Integer.valueOf(bs[i]);
        }

        int[] box = new int[boxes];
        for (int i = 0; i < box.length; i++) {
            box[i] = 0;
        }

        for (int i = bells.length - 1; i >= 0; i--) {
            for (int j = 0; j < box.length; j++, i--) {
                if (i >= 0) {
                    box[j] += bells[i];
                } else {
                    break;
                }
            }
            for (int j = box.length - 1; j >= 0; j--, i--) {
                if (i >= 0) {
                    box[j] += bells[i];
                } else {
                    break;
                }
            }
        }

        //get max
        int max = Integer.MIN_VALUE;
        for (int i = 0; i < box.length; i++) {
            if (box[i] > max) {
                max = box[i];
            }
        }

        System.out.println(max);

        s.close();
    }
}
