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.Arrays;
import java.util.StringTokenizer;


public class B_Div2_334 {
    int n;
    int[]s;
    public static void main(String[]arg) throws IOException
    {
        new B_Div2_334().solve();
    }
    public void solve() throws IOException
    {
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(in.readLine());
        
        int k,i,j;
        n = Integer.parseInt(st.nextToken());
        k = Integer.parseInt(st.nextToken());
        s = new int[n];
        st = new StringTokenizer(in.readLine());
        int sum = 0;
        for(i = 0; i < n; i++)
            s[i] = Integer.parseInt(st.nextToken());
        int right = 10000000+n, left = 1,middle;
        int kp;
        while(left < right)
        {
            middle = (left+right)/2;
            kp = numBoxs(middle);
            if(kp > k)
                left = middle + 1;
            else
                right = middle;
        }
        System.out.println(left);
    }
    private int numBoxs(int cap)
    {
        int i = 0, j = n -1;
        int bs = 0;
        boolean end = false;
        while(i <= j && !end)
        {
            if(s[i]+s[j] > cap)
            {
                if(s[j] > cap)
                    end = true;
                j--;
            }
            else
            {
                i++;j--;
            }
            bs++;
        }
        if(end) bs = 1000000;
        return bs;
    }
}