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 GCD(int a, int b) { return a == 0 ? b : GCD(b % a, a); }
string ntoa(int b) {
  string a = "";
  while (b) {
    a += (b % 10 + '0');
    b /= 10;
  }
  reverse(a.begin(), a.end());
  return a;
}
struct myitem {
  int h;
  int w;
  myitem(int a, int b) {
    h = a;
    w = b;
    if (h > w) {
      int t = h;
      h = w;
      w = t;
    }
  }
};
bool mycompare(myitem a, myitem b) {
  if (a.h == b.h) return a.w < b.w;
  return a.h < b.h;
}
int main() {
  int n, k;
  cin >> n >> k;
  int mxbox = -1;
  vector<int> bellsize;
  for (int i = 0; i < n; ++i) {
    int a;
    cin >> a;
    bellsize.push_back(a);
  }
  int j = n - 1;
  int left = n;
  while ((k * 2) != left && left) {
    if (bellsize[j] > mxbox) {
      mxbox = bellsize[j];
    }
    k--;
    left--;
    j--;
  }
  int i = 0;
  while (left) {
    if (bellsize[i] + bellsize[j] > mxbox) {
      mxbox = bellsize[i] + bellsize[j];
    }
    i++;
    j--;
    left -= 2;
  }
  cout << mxbox << "\n";
}
