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.*;

public class Main {

    public static void main(String[] args) throws IOException {
        Reader.init(System.in);
//        ArrayList<Integer>a=new ArrayList<>();
//        a.add(2);a.add(5);a.add(6);a.add(10);
//        System.out.println(BinarySearch(a, 3));
        int n = Reader.nextInt(),k=Reader.nextInt();
        ArrayList<Integer>a = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            a.add(Reader.nextInt());
        }
        int max = a.get(n-1);
      
        int d=(2*k)-n;
        while(d-->0&&a.size()>0){
            //System.out.println(a.get(a.size()-1));
            k--;
            a.remove(a.size()-1);}
        for (int i = 0,j=a.size()-1; i < a.size(); i++,j--) {
            max=Integer.max(a.get(i)+a.get(j),max);
        }
         System.out.println(max);
    }

    static int BinarySearch(ArrayList<Integer>a, int key) {
        int l = 0, h = a.size();
        while (l < h) {
            int mid = (l + h) / 2;
            if (key >= a.get(mid)) {
                l = mid+1;
            } else {
                h = mid;
            }

        }
        return l-1;
    }
}

class Reader {

    static StringTokenizer tokenizer;
    static BufferedReader reader;

    public static void init(InputStream input) throws UnsupportedEncodingException {
        reader = new BufferedReader(new InputStreamReader(input, "UTF-8"));
        tokenizer = new StringTokenizer("");
    }

    public static String next() throws IOException {
        while (!tokenizer.hasMoreTokens()) {
            tokenizer = new StringTokenizer(reader.readLine());
        }
        return tokenizer.nextToken();
    }

    public static String nextLine() throws IOException {
        return reader.readLine();
    }

    public static int nextInt() throws IOException {
        return Integer.parseInt(next());
    }

    public static double nextDouble() throws IOException {
        return Double.parseDouble(next());
    }

    public static long nextLong() throws IOException {
        return Long.parseLong(next());
    }
}
