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}.
/**
 * Created by 1 on 16.12.2015.
 */

import java.util.*;
import java.io.*;

public class Main {
    public static int a[];
    public static int k;
    public static boolean f(int s){
        int ans = 0;
        int indl = 0;
        for(int i = a.length - 1;i >= 0;i--) {
            if (i >= indl) {
                if (i != indl) {
                    if (a[i] + a[indl] <= s) {
                        ans++;
                        indl++;
                    } else if (a[i] <= s)
                        ans++;
                    else
                        return false;
                } else {
                    if (a[i] <= s)
                        ans++;
                    else
                        return false;
                    break;
                }
            }
        }
        return (ans <= k);
    }
    public static void main(String argc[]){
        Scanner in = new Scanner(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int n;
        n = in.nextInt();
        k = in.nextInt();

        a = new int[n];
        for(int i = 0;i < n;i++)
            a[i] = in.nextInt();

        int l = 0 , r = 2000001;
        while(r - l != 1){
            int mid = (l + r) / 2;
            if(f( mid ))
                r = mid;
            else
                l = mid;
        }
        out.print(r);
        out.close();
    }
}
