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 int arr[1000001], n, k;
bool solve(long long int m) {
  if (k == n) {
    if (m >= arr[n - 1]) {
      return true;
    } else {
      return false;
    }
  } else if (k == n / 2) {
    long long int f = 0, j = n;
    for (long long int i = 1; i <= n / 2; i++) {
      f = max(f, arr[i] + arr[j]);
      j--;
    }
    if (m >= f) {
      return true;
    } else {
      return false;
    }
  } else {
    long long int e = (k * 2) - n;
    long long int f = 0;
    for (long long int j = n; j > n - e; j--) {
      f = max(f, arr[j]);
    }
    long long int j = n - e;
    for (long long int i = 1; i <= (n - e) / 2; i++) {
      f = max(f, arr[i] + arr[j]);
      j--;
    }
    if (m >= f) {
      return true;
    } else {
      return false;
    }
  }
}
long long int solvebs() {
  long long int l = arr[n] - 1, h = arr[n] + arr[n - 1];
  long long int m;
  while (h - l > 1) {
    m = (l + h) / 2;
    if (solve(m)) {
      h = m;
    } else {
      l = m;
    }
  }
  return h;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  t = 1;
  while (t--) {
    cin >> n >> k;
    for (long long int i = 1; i <= n; i++) {
      cin >> arr[i];
    }
    if (k > n) {
      cout << arr[n];
    } else {
      long long int ans;
      ans = solvebs();
      cout << ans;
    }
  }
  return 0;
}
