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 = 100010;
const int M = 1000010;
const int MOD = 1000000007;
const long long INF = 0x7fffffff;
const int dir4[4][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
const int dir8[8][2] = {{-1, 0}, {-1, 1}, {0, 1},  {1, 1},
                        {1, 0},  {1, -1}, {0, -1}, {-1, -1}};
const double eps = 1e-8;
const double PI = acos(-1.0);
inline int sign(double x) { return (x > eps) - (x < -eps); }
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
inline T Min(T a, T b) {
  return a < b ? a : b;
}
template <class T>
inline T Max(T a, T b) {
  return a > b ? a : b;
}
int n, k, f[N];
int main() {
  while (scanf("%d%d", &(n), &(k)) != EOF) {
    int mx = -1;
    for (int i = 0; i < (n); i++) {
      scanf("%d", &(f[i]));
      mx = max(mx, f[i]);
    }
    if (k >= n) {
      printf("%d\n", (mx));
      continue;
    }
    int i = n - k - 1, t = n - k;
    while (i >= 0) {
      f[t] += f[i];
      t++;
      i--;
    }
    for (int i = n - k; i < n; i++) mx = max(mx, f[i]);
    printf("%d\n", (mx));
  }
  return 0;
}
