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;
import java.util.StringTokenizer;

public class R334Div2B {

	public static void main(String[] args) {
		FastScanner in=new FastScanner();
		int n=in.nextInt();
		int k=in.nextInt();
		int[] s=new int[n];
		long hi=0;
		for(int i=0;i<n;i++){
			s[i]=in.nextInt();
			hi+=s[i];
		}
		
		if(k>=n){
			System.out.println(s[n-1]);
			return;
		}
		
		int dbl=n-k;
		//int sgl=n-dbl*2;
		
		long lo=1,mid;
		while(lo<hi){
			mid=lo+(hi-lo)/2;
			long max=s[n-1];
			for(int i=0;i<dbl;i++){
				max=Math.max(max,s[i]+s[dbl*2-i-1]);
			}
			if(max<=mid){
				hi=mid;
			}else{
				lo=mid+1;
			}
		}
		
		System.out.println(lo);
	}
	static class FastScanner{
		BufferedReader br;
		StringTokenizer st;
		public FastScanner(){br=new BufferedReader(new InputStreamReader(System.in));}
		String nextToken(){
			while(st==null||!st.hasMoreElements())
				try{st=new StringTokenizer(br.readLine());}catch(Exception e){}
			return st.nextToken();
		}
		int nextInt(){return Integer.parseInt(nextToken());}
		long nextLong(){return Long.parseLong(nextToken());}
		double nextDouble(){return Double.parseDouble(nextToken());}
	}

}
