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 File {
    public static class FastScanner {
        BufferedReader br;
        StringTokenizer st;

        public FastScanner() {
            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());
        }
    }
	
    public static void main(String[] args) {
        FastScanner sc = new FastScanner();
        PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
		
		
	
		int n = sc.nextInt();
		int k = sc.nextInt();
		
		int[] s = new int[n];
		for (int i = 0; i < n; i++) {
			s[i] = sc.nextInt();
		}
		
		
		
		long lo = s[n-1];
		long hi = s[n-1] * 2;
		long res = hi;
		
		
		while (lo <= hi) {
			long mid = lo + (hi - lo) / 2;
			
			if (isPossible(s, mid, k)) {
				res = Math.min(res, mid);
				hi = mid - 1;
			}
			else {
				lo = mid + 1;
			}
		}
		
		
		out.println(res);
		
		
		
		out.close();
	}
	
	public static boolean isPossible(int[] a, long S, long k) {
		int boxes = 0;
		
		int lo = 0;
		int hi = a.length-1;
		
		while (lo <= hi) {
			boxes++;
			
			long curr = S;
			curr -= a[hi--];
			
			if (lo <= hi && a[lo] <= curr) {
				lo++;
			}
		}
		
		return boxes <= k;
	}
}












	