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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;
import static java.util.Arrays.sort;
import javafx.util.Pair;

public class Main
{

    public static void main(String[] args)
    {

        FastScanner input = new FastScanner();
        int n = input.nextInt();
        int k = input.nextInt();
        int a[] = new int[n];
    
        for (int i = 0; i < n; i++) {
            a[i] = input.nextInt();
            
            
        } 
        if(n<=k)
        {
            System.out.println(a[n-1]);
        }
        else if(n==2*k)
        {
            int i =0;
            int j =n-1;
            int max= Integer.MIN_VALUE;
            while(i<=j)
            {
                max = Math.max(max, a[i]+a[j]);
                i++;
                j--;
            }
            System.out.println(max);
        }
        else
        {
            int i = 0;
            int j = n-(k*2-n)-1;
            int max= Integer.MIN_VALUE;
            while(i<=j)
            {
                max = Math.max(max, a[i]+a[j]);
                i++;
                j--;
            }
            max = Math.max(max, a[a.length-1]);
            System.out.println(max);
            
        }

    }

    static class FastScanner
    {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer("");

        String next()
        {
            while (!st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {

            return Long.parseLong(next());
        }

        double nextDouble()
        {
            return Double.parseDouble(next());
        }

        String nextLine() throws IOException
        {
            return br.readLine();
        }
    }

}
