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.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;

public class test {
	private static void scores(int[] ss, int[] ms, int[] ws, int[] hs) {
		int sum = 0;
		for(int i=0; i<5; i++)
			sum += Math.max(ss[i]*3/10, ss[i]-ss[i]/250*ms[i]-50*ws[i]);
		sum += 100*hs[0] - 50*hs[1];
		System.out.println(sum);
	}
	
	private static int needBox(int[] cows, int size) {
		int l=0, r=cows.length-1, count=0;
		while(l < r) {
			count++;
			if(cows[l]+cows[r] <= size) {
				l++;
				r--;
			} else {
				r--;
			}
		}
		if(l == r)  count++;
		return count;
	}
	
	private static void smallestBox(int[] cows, int k) {
		if(cows.length == 1) {
			System.out.println(cows[0]);
			return;
		}
		
		int n = cows.length, min = cows[n-1], max = cows[n-1]+cows[n-2];
		while(min < max) {
			int mid  = (max-min)/2+min;
			int need = needBox(cows, mid);
			if(need > k)  min = mid+1;
			else          max = mid;
		}
		System.out.println(min);
	}
	
	public static void main(String[] args) {
		//new careercup().run();
		try{
			BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
			String[] nk = br.readLine().split(" ");
			int n = Integer.parseInt(nk[0]);
			int k = Integer.parseInt(nk[1]);
			String[] ss = br.readLine().split(" ");
			int[] cows = new int[n];
			for(int i=0; i<n; i++)
				cows[i] = Integer.parseInt(ss[i]);
			smallestBox(cows, k);
		}catch(IOException io){
			io.printStackTrace();
		}
	}
}
