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.*;
import java.util.*;


public class Fight {
	static Scanner sc=new Scanner(System.in);
	static PrintWriter out=new PrintWriter(System.out);
	private static ArrayList<Integer> primes;
	
	public static void main(String[] args) throws IOException {	
		int n=sc.nextInt();
		int k=sc.nextInt();
		int a[]=new int[n];
		long sum=0;
		for (int i = 0; i < a.length; i++) {
			a[i]=sc.nextInt();
			sum+=a[i];
		}
		long high=a[n-1]*2;
		long low=a[n-1];
		long ans=0;
		while(high>=low) {
			long mid=(high+low)/2;
			if(works(mid,a,k,sum)) {
				high=mid-1;
				ans=mid;
			}
			else {
				
				low=mid+1;
			}
		}
		System.out.println(ans);


		
	out.flush();	
	}
	
	
	private static boolean works(long mid,int[] a,int m,long sum) {
		int i, j;
		for(i = 0, j = a.length - 1; m> 0 && i <= j;m--,i++,j--) {
			if(a[i] + a[j] >mid)
			{
			i--;
			}
	}
		return i>j;
	
	}


	static void sieve(int N)	// O(N log log N) 
	{
		int[] isComposite = new int[N+1];					
		isComposite[0] = isComposite[1] = 1;			// 0 indicates a prime number
		primes = new ArrayList<Integer>();

		for (int i = 2; i <= N; ++i) 					//can loop till i*i <= N if primes array is not needed O(N log log sqrt(N)) 
			if (isComposite[i] == 0) 					//can loop in 2 and odd integers for slightly better performance
			{
				
				primes.add(i);
				if(1l * i * i <= N)
					for (int j = i * i; j <= N; j += i)	// j = i * 2 will not affect performance too much, may alter in modified sieve
						isComposite[j] = 1;
			}   
	}static  class Scanner {
	StringTokenizer st;
	BufferedReader br;

	public Scanner(InputStream s) {
		br = new BufferedReader(new InputStreamReader(s));
	}

	public String next() throws IOException {
		while (st == null || !st.hasMoreTokens())
			st = new StringTokenizer(br.readLine());
		return st.nextToken();
	}

	public int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	public int[] nextIntArr(int n) throws IOException {
		int[] arr = new int[n];
		for (int i = 0; i < n; i++)
			arr[i] = nextInt();
		return arr;
	}

	public long[] nextLongArr(int n) throws IOException {
		long[] arr = new long[n];
		for (int i = 0; i < n; i++)
			arr[i] = nextLong();
		return arr;
	}

	public long nextLong() throws IOException {
		return Long.parseLong(next());
	}

	public String nextLine() throws IOException {
		return br.readLine();
	}

	public double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}

	public boolean ready() throws IOException {
		return br.ready();
	}
}
}

