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.io.*;
import java.util.*;
import java.lang.String;
import java.util.Arrays;

import static java.lang.Integer.*;

public class Main extends PrintWriter {
    Main() { super(System.out, true); }
    Scanner sc = new Scanner(System.in);
    public static void main(String[] $) {
        Main o = new Main(); o.main(); o.flush();
    }
    //took bsearch from Dukkha
    private static int binSearch(Integer[] arr, int number){
        int left=0,right=arr.length-1,mid=(left+right)/2,ind=0;
        while(left<=right){
            if(arr[mid]<=number){
                ind=mid+1;
                left=mid+1;
            }
            else
                right=mid-1;
            mid=(left+right)/2;
        }
        return ind;
    }
//used tutoral to help understand problem
    void main() {
        int res = 0;
        int N = sc.nextInt();
        int K = sc.nextInt();
        Integer [] S = new Integer [N];
        for(int i = 0; i < N; i++){
            S[i] = sc.nextInt();
            res = max(res, S[i]);
        }
        for(int i = 0; i < N - K; i++){
            res = max(res, S[i] + S[2 * (N - K) - 1 - i]);
        }
        print(res);
    }
}