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;
long long n, m, k, a[100001];
template <typename T>
inline T sqr(T a) {
  return a * a;
}
template <typename T>
inline T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
inline bool Palindrome(const string& s) {
  return equal(s.begin(), s.end(), s.rbegin());
}
inline bool check(int x) {
  int l = 1, r = n, cnt = 0;
  while (l < r) {
    if (a[r] > x || a[l] > x) {
      return 0;
    }
    if (a[l] + a[r] <= x) {
      l++;
      r--;
    } else
      r--;
    cnt++;
  }
  if (a[1] > x) {
    return 0;
  }
  if (l == r) {
    cnt++;
  }
  return cnt <= k;
}
int main() {
  int ans = 1e9;
  cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int l = 1, r = 1e9;
  while (l <= r) {
    int mid = (l + r) / 2;
    if (check(mid)) {
      ans = min(mid, ans);
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  cout << ans;
}
