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.*;
import java.math.*;
 
public class Main {
 
    FastScanner in;
    PrintWriter out;
 
    static class FastScanner {
 
        BufferedReader br;
        StringTokenizer st;
 
        FastScanner(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
            st = null;
        }
 
        String next() throws IOException {
            while (st == null || !st.hasMoreTokens())
                st = new StringTokenizer(br.readLine());
            return st.nextToken();
        }
 
        String nextLine() throws IOException {
            if (st == null || !st.hasMoreTokens())
                return br.readLine();
            StringBuilder result = new StringBuilder(st.nextToken());
            while (st.hasMoreTokens()) {
                result.append(" ");
                result.append(st.nextToken());
            }
            return result.toString();
        }
 
        int nextInt() throws IOException {
            return Integer.parseInt(next());
        }
 
        long nextLong() throws IOException {
            return Long.parseLong(next());
        }
 
        double nextDouble() throws IOException {
            return Double.parseDouble(next());
        }
 
    }
 
    void run() throws IOException {
        in = new FastScanner(System.in);
        out = new PrintWriter(System.out, false);
        solve();
        out.close();
    }
 
    public static void main(String[] args) throws IOException{
        new Main().run();
    }
 
    public void printArr(int[] arr){
        for(int i = 0; i < arr.length; i++){
            out.print(arr[i] + " ");
        }
        out.println();
    }
 
    public long gcd(long a, long b){
        if(a == 0) return b;
        return gcd(b % a, a);
    }

    int n;
    int k;
    int[] ss;
    int res = 1000000000;

    public void solve() throws IOException{

        n = in.nextInt(); 
        k = in.nextInt();
        ss = new int[n];
        
        for(int i = 0; i < n; i++){
            int num = in.nextInt();
            ss[i] = num;
        }

        binarySearch(1, 10000001);
        out.println(res);
        return;
    }

    public void binarySearch(int l, int r){

        int mid = (l + r) / 2;
        if(l == r) return;
        if(check(mid)){
            res = Math.min(res, mid);
            binarySearch(l, mid);
        }else{
            binarySearch(mid + 1, r);
        }

    }

    public boolean check(int s){

        int t = 0, l = 0, r = n;

        while( l < r ){
            if(l == r - 1){
                if(ss[r - 1] <= s){
                    t++;
                    l++;
                }else{
                    return false;
                }
            }else if(ss[r - 1] + ss[l] <= s){
                l++; r--; t++;
            }else if(ss[r - 1] <= s){
                r--; t++;
            }else{
                return false;
            }
        }

        return t <= k;

    }

}