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.math.*;
import java.util.*;
 
public class Main {
	static final long MOD = 998244353;
	//static final long MOD = 1000000007;
	static boolean[] visited;
    public static void main(String[] args) throws IOException {
        FastScanner sc = new FastScanner();
        int N = sc.nextInt();
        int K = sc.nextInt();
        int[] sizes = new int[N];
        for (int i = 0 ; i< N; i++)
        	sizes[i] = sc.nextInt();
        
        int low = sizes[N-1];
        int high = 2*sizes[N-1];
        while (low < high) {
        	int mid = (low+high)/2;
        	int boxesUsed = 0;
        	int size = 0;
        	TreeMap<Integer,Integer> tm = new TreeMap<Integer,Integer>();
        	for (int i = N-1; i >= 0; i--) {
        		int rem = mid - sizes[i];
        		//Add this bell into a pending box and then remove the pending box.
        		if (tm.floorKey(rem) != null) {
        			int key = tm.floorKey(rem);
        			if (tm.get(key) == 1)
        				tm.remove(key);
        			else
        				tm.put(key, tm.get(key)-1);
        		} else {
        			//Make this box a pending box
        			boxesUsed++;
        			tm.put(sizes[i],tm.getOrDefault(sizes[i],0)+1);
        		}
        	}
    		if (boxesUsed <= K) {
    			high = mid;
    		} else {
    			low = mid+1;
    		}
        }
        System.out.println(low);
    }
    
    static class FastScanner { 
        BufferedReader br; 
        StringTokenizer st; 
  
        public FastScanner() { 
            br = new BufferedReader(new InputStreamReader(System.in)); 
        } 
  
        String next() { 
            while (st == null || !st.hasMoreElements()) { 
                try { 
                    st = new StringTokenizer(br.readLine());
                } catch (IOException  e) { 
                    e.printStackTrace(); 
                } 
            } 
            return st.nextToken(); 
        } 
  
        int nextInt() { 
            return Integer.parseInt(next()); 
        } 
  
        long nextLong() { 
            return Long.parseLong(next()); 
        } 
  
        double nextDouble() { 
            return Double.parseDouble(next()); 
        } 
  
        String nextLine() { 
            String str = ""; 
            try { 
                str = br.readLine(); 
            } catch (IOException e) {
                e.printStackTrace(); 
            } 
            return str; 
        }
    }
}