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 main() {
  int n, k;
  cin >> n >> k;
  long long int a[n];
  long long int high = 0, low = 0;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    high += a[i];
  }
  low = a[0];
  while (low < high) {
    long long int mid = (low + high) / 2;
    int f = 0;
    long long int boxes[k];
    memset(boxes, 0, sizeof(boxes));
    int i = n - 1;
    int ctr = 0;
    while (ctr < k) {
      if (ctr == n) {
        f = 1;
        break;
      }
      if (a[i] > mid) {
        f = 2;
        break;
      }
      boxes[ctr] = a[i];
      i--;
      ctr++;
    }
    if (f == 0) {
      int x = 0;
      f = 1;
      for (int j = 0; j <= i; j++) {
        int placed = 0;
        while (x < k) {
          if (x == k - 1) {
            if (boxes[x] + a[j] > mid) {
              f = 2;
              break;
            }
          }
          if (boxes[x] + a[j] <= mid) {
            boxes[x] += a[j];
            x++;
            placed = 1;
            break;
          } else
            x++;
        }
        if (placed == 0) f = 2;
        if (f == 2) break;
      }
    }
    if (f == 2) {
      low = mid + 1;
    } else if (f == 1)
      high = mid;
  }
  cout << high;
}
