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;
const int N = 2e5 + 5;
int n, k;
long long arr[N];
long long sum;
deque<long long> deq;
vector<pair<long long, long long> > vec;
int main() {
  scanf("%d%d", &n, &k);
  vec.resize(k);
  for (int i = 0; i < n; i++)
    scanf("%I64d", arr + i), sum += arr[i], deq.push_back(arr[i]);
  if (k == 1)
    printf("%I64d\n", sum);
  else if (n <= k)
    printf("%I64d\n", arr[n - 1]);
  else if (n == 1)
    printf("%I64d\n", arr[0]);
  else {
    long long mx = arr[n - 1];
    deq.pop_back();
    vec.push_back(make_pair(mx, 0));
    for (int i = 1; deq.size() && i < k; i++) {
      vec.push_back(make_pair(deq.back(), 0));
      deq.pop_back();
    }
    while (deq.size() && vec.size()) {
      vec.back().second = deq.back();
      mx = max(mx, vec.back().first + vec.back().second);
      vec.pop_back();
      deq.pop_back();
    }
    for (int i = 0; i < vec.size(); i++)
      mx = max(mx, vec[i].first + vec[i].second);
    printf("%I64d\n", mx);
  }
  return 0;
}
