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.Scanner;

public class Main {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int n = sc.nextInt();
		int k = sc.nextInt();
		int[] s = new int[n];
		for (int i = 0; i < s.length; i++) {
			s[i] = sc.nextInt();
		}
		System.out.println(solve(s, k));

		sc.close();
	}

	static int solve(int[] s, int k) {
		int result = -1;
		int lower = s[s.length - 1];
		int upper = s[s.length - 1] * 2;
		while (lower <= upper) {
			int middle = (lower + upper) / 2;

			if (computeBoxNum(s, middle) <= k) {
				result = middle;

				upper = middle - 1;
			} else {
				lower = middle + 1;
			}
		}

		return result;
	}

	static int computeBoxNum(int[] s, int boxSize) {
		int boxNum = 0;
		int freeCount = 0;
		int leftIndex = 0;
		int rightIndex = s.length - 1;
		while (leftIndex <= rightIndex) {
			while (leftIndex != rightIndex && s[leftIndex] + s[rightIndex] <= boxSize) {
				leftIndex++;
				freeCount++;
			}

			if (freeCount != 0) {
				freeCount--;
			}
			boxNum++;

			rightIndex--;
		}

		boxNum += (freeCount + 1) / 2;

		return boxNum;
	}
}
