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 solution;

import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.PriorityQueue;
import java.util.StringTokenizer;
import java.util.TreeSet;

public class Solution implements Runnable {
    
    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok = new StringTokenizer("");
    final boolean OJ = System.getProperty("ONLINE_JUDGE") != null;
    
    void init() throws FileNotFoundException {
        if (OJ) {
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
        } else {
            in = new BufferedReader(new FileReader("input.txt"));
            out = new PrintWriter("output.txt");
        }
            
    }
    
    public void run() {
        try {
            init();
            solve();
            out.close();
        } catch (Exception e) {
            e.printStackTrace();
            System.exit(-1);
        }
    }
    
    String readString() throws IOException {
        while (!tok.hasMoreTokens()) {
            tok = new StringTokenizer(in.readLine());
        }
        
        return tok.nextToken();
    }
    
    int readInt() throws IOException {
        return Integer.parseInt(readString());
    }
    
    long readLong() throws IOException {
        return Long.parseLong(readString());
    }

    public static void main(String[] args) {
        new Thread(null, new Solution(), "", 256 * 1l << 20).start();
    }
    
    int[] a;
    int n;
    int k;
    
    
    void solve() throws IOException {
        n = readInt();
        k = readInt();
        a = new int[n];
        int ans;
        for (int i = 0; i < n; i++) {
            a[i] = readInt();
        }
        ans = a[n - 1];
        for (int i = 0; i < n - k; i++) {
            ans = Math.max(ans, a[n - k + i] + a[n - k - 1 - i]);
        }
        out.print(ans);
    }
}
