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>
template <class T>
bool valid(T x, T w) {
  return 0 <= x && x < w;
}
long long di[4] = {1, -1, 0, 0};
long long dj[4] = {0, 0, 1, -1};
using namespace std;
long long score[5] = {(long long)(500 * 0.3), (long long)(1000 * 0.3),
                      (long long)(1500 * 0.3), (long long)(2000 * 0.3),
                      (long long)(2500 * 0.3)};
long long divscore[5] = {2, 4, 6, 8, 10};
bool f(deque<long long> bells, long long nbox, long long size) {
  nbox--;
  while (!bells.empty()) {
    if (nbox < 0) {
      return false;
    }
    auto x = bells.front();
    bells.pop_front();
    if (bells.size() > 0) {
      auto y = bells.back();
      if (x + y <= size) {
        bells.pop_back();
      }
      nbox--;
      continue;
    }
  }
  return true;
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  long long n, k;
  deque<long long> s;
  cin >> n >> k;
  for (long long(i) = 0; (i) < (n); ++(i)) {
    long long c;
    cin >> c;
    s.push_front(c);
  }
  long long left, right, middle;
  left = s[0] - 1;
  right = s[0] * 2 + 1;
  while (right - left > 1) {
    middle = (left + right) / 2;
    (f(s, k, middle) ? right : left) = middle;
  }
  cout << right << endl;
}
