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}.
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k, arr[100001];
  vector<int> vec;
  cin >> n >> k;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
  }
  if (k >= n) {
    cout << arr[n - 1];
    return 0;
  }
  int ans1 = 0;
  while (true) {
    if ((n - ans1) / 2 + ans1 == k) break;
    ans1++;
  }
  int x = n - ans1;
  x /= 2;
  x *= 2;
  for (int i = 0; i < x / 2; i++) {
    vec.push_back(arr[i] + arr[x - i - 1]);
  }
  for (int i = x; i < n; i++) {
    vec.push_back(arr[i]);
  }
  sort(vec.begin(), vec.end());
  cout << vec[vec.size() - 1];
  return 0;
}
