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 int fitK(int[] boxes, int k) {
        int l = boxes.length; 
        if (l == 0) return 0; 
        if (l == 1) return boxes[0]; 
        if (l == k) return boxes[l - 1];
        if (l == 2 * k) return findMax2(boxes, l); 
        int end = 2 * (l - k); 
        return Math.max(findMax2(boxes, end), boxes[l - 1]); 
    }
    
    public static int findMax2(int[] boxes, int end) {
        int max = 0; 
        for (int i = 0; i <= end / 2; i++) {
            if (boxes[i] + boxes[end - 1 - i] > max) max = boxes[i] + boxes[end - i - 1]; 
        }
        return max; 
    }
    
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in); 
		int length = in.nextInt(); 
		int k = in.nextInt(); 
		int[] input = new int[length]; 
		for (int i = 0; i < length; i++) {
			input[i] = in.nextInt(); 
		}
		System.out.println(fitK(input, k));
	}

}
