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}.
//package codeforces;

import java.io.*;
import java.util.*;
 
 
public class TheatreSquare{
   public static void main(String[] args) {
      MyScanner sc = new MyScanner();
      out = new PrintWriter(new BufferedOutputStream(System.out));
      
      // Start writing your solution here. -------------------------------------
   
      /*
      int n      = sc.nextInt();        // read input as integer
      long k     = sc.nextLong();       // read input as long
      double d   = sc.nextDouble();     // read input as double
      String str = sc.next();           // read input as String
      String s   = sc.nextLine();       // read whole line as String

      int result = 3*n;
      out.println(result);                    // print via PrintWriter
      */
      int n = sc.nextInt();
      int k = sc.nextInt();
      int[] arr = new int[n];
      for(int j=0; j<n; j++){
    	  arr[j] = sc.nextInt();;
      }
      int res = arr[n-1];
      
      for(int i=0; i<n-k; i++){ 
    	 res = Math.max(res, arr[i]+arr[2*n-2*k-i-1]);
      }
      
      out.println(res);  
      // Stop writing your solution here. -------------------------------------
      out.close();
   }
   
 //-----------PrintWriter for faster output---------------------------------
   public static PrintWriter out;
      
   //-----------MyScanner class for faster input----------
   public static class MyScanner {
      BufferedReader br;
      StringTokenizer st;
 
      public MyScanner() {
         br = new BufferedReader(new InputStreamReader(System.in));
      }
 
      String next() {
          while (st == null || !st.hasMoreElements()) {
              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(){
          String str = "";
	  try {
	     str = br.readLine();
	  } catch (IOException e) {
	     e.printStackTrace();
	  }
	  return str;
      }

   }
   //--------------------------------------------------------
}
