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 dx8[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dy8[] = {-1, 1, 0, 0, -1, -1, 1, 1};
int dy[] = {1, -1, 0, 0};
int dx[] = {0, 0, 1, -1};
int X[6] = {0, 0, 0, -1, 0, 1};
int Y[6] = {0, 0, -1, 0, 1, 0};
int Z[6] = {1, -1, 0, 0, 0, 0};
int n, k;
vector<int> v;
int boxes = 0;
bool check(long long mid) {
  boxes = 0;
  for (int i = 0, j = n - 1; i <= j;) {
    if (v[i] + v[j] <= mid && i != j) {
      boxes++;
      i++, j--;
    } else if (v[j] > v[i] && v[j] <= mid) {
      boxes++;
      j--;
    } else if (v[i] >= v[j] && v[i] <= mid) {
      boxes++;
      i++;
    } else
      return 0;
  }
  return 1;
}
int main() {
  cin >> n >> k;
  v.resize(n);
  for (int i = 0; i < n; i++) scanf("%d", &v[i]);
  long long s = 1, e = 1e12, mid, ans = 1e12;
  while (s <= e) {
    mid = s + (e - s) / 2;
    if (check(mid) && boxes <= k) {
      ans = mid;
      e = mid - 1;
    } else
      s = mid + 1;
  }
  cout << ans << "\n";
}
