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.io.BufferedReader;
import java.io.InputStreamReader;

public class Div334B {

    public static void main(String[] args) throws Exception {
        BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
        String[] l = bf.readLine().split(" ");
        int n = Integer.parseInt(l[0]);
        int k = Integer.parseInt(l[1]);
        int[] nums = new int[n];
        l = bf.readLine().split(" ");
        int max = 0;
        for (int i = 0; i < nums.length; i++) {
            nums[i] = Integer.parseInt(l[i]);
            max = Math.max(nums[i], max);
        }
        long lo = 0;
        long mid, hi = 10000000;
        while (lo + 1 < hi) {
            mid = (lo + hi) / 2;
            if (check(mid, k, nums) || check2(mid, k, nums)) {
                hi = mid;
            } else {
                lo = mid;
            }
        }
        System.out.println(hi);
    }

    private static boolean check2(long mid, int k, int[] nums) {
        int n = nums.length;
        int x = 0;
        for (int i = n - 1; i >= x; i--) {
            if (k == 0 || nums[i] > mid)
                return false;
            if (x < i && nums[i] + nums[x] <= mid) {
                x++;
            }
            k--;
        }
        return true;
    }

    private static boolean check(long mid, int k, int[] nums) {
        int n = nums.length;
        for (int i = 0; i < n; i++) {
            if (k == 0 || nums[i] > mid)
                return false;
            if (i < n - 1 && nums[i] + nums[i + 1] <= mid) {
                i++;
            }
            k--;
        }
        return true;
    }
}