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 B {
    public static void main(String[] args) {

        QuickScanner in = new QuickScanner(System.in);
        int n = in.nextInt();
        int k = in.nextInt();
        int[] s = new int[n];
        for (int i = 0; i < n; i++)
            s[i] = in.nextInt();
        if (n <= k)
        {
            System.out.println(s[n-1]);
            return;
        }
        int res = Integer.MIN_VALUE;
        int offs = 0;
        for (int kt = k; kt * 2 > n - offs; kt--) offs++;
        for (int i = 0; i < (n-offs) / 2; i++)
        {
            int j = (n-offs) - 1 - i;
            res = Math.max(res, s[i] + s[j]);
        }
        if (2 * k > n) System.out.println(Math.max(s[n-1], res));
        else System.out.println(res);
    }
}



































class QuickScanner {
    BufferedReader in;
    StringTokenizer token;
    String delim;
    public QuickScanner(InputStream inputStream)
    {
        this.in=new BufferedReader(new InputStreamReader(inputStream));
        this.delim=" \n\t";
        this.token=new StringTokenizer("",delim);
    }

    public boolean hasNext()
    {
        while(!token.hasMoreTokens())
        {
            try {
                String s=in.readLine();
                if(s==null) return false;
                token=new StringTokenizer(s,delim);
            } catch (IOException e) {
                throw new InputMismatchException();
            }
        }
        return true;
    }

    public String next()
    {
        hasNext();
        return token.nextToken();
    }

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

}
