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

public class CF604B{ 
  public static void main(String[] args)throws IOException, Exception {
    InputReader in = new InputReader(System.in);
    PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
    int n = in.nextInt(), k = in.nextInt();
    int [] a = new int[n];
    for(int i = 0; i < n; i++){
      a[i] = in.nextInt();
    }
    int two =Math.max(0, n - k);
    //if(k > n) two = 0;
   // System.err.println(two);
    int max = 0;
    for(int i = 0, j = 2*two-1; i < two ; i++, j--){
      max = Math.max(max, a[i]+a[j]);
    }
    for(int i = 2*two; i < n; i++){
     // System.err.println(i);
      max = Math.max(max, a[i]);
    }
  
    
    pw.println(max);
    pw.close();
  }
  
  
  static class InputReader {
    public BufferedReader reader;
    public StringTokenizer tokenizer;
    
    public InputReader(InputStream stream) {
      reader = new BufferedReader(new InputStreamReader(stream));
      tokenizer = null;
    }
    
    public String next() {
      try {
        while (tokenizer == null || !tokenizer.hasMoreTokens())
          tokenizer = new StringTokenizer(reader.readLine());
      } catch (IOException e) {
        throw new RuntimeException(e);
      }
      
      return tokenizer.nextToken();
    }
    
    public String nextLine() {
      String line = null;
      try {
        tokenizer = null;
        line =  reader.readLine();
      } catch (IOException e) {
        throw new RuntimeException(e);
      }
      return line;
    }
    
    public int nextInt() {
      return Integer.parseInt(next());
    }
    
    public double nextDouble() {
      return Double.parseDouble(next());
    }
    
    public long nextLong() {
      return Long.parseLong(next());
    }
  }
}