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>
long long int pow(long long int a, long long int b, long long int m) {
  long long int x = 1, y = a;
  while (b > 0) {
    if (b % 2 == 1) {
      x = (x * y);
      if (x > m) x %= m;
    }
    y = (y * y);
    if (y > m) y %= m;
    b /= 2;
  }
  return x % m;
}
long long int gcd(long long int a, long long int b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
using namespace std;
long long int a[100001];
long long int b[100001];
int main() {
  ios_base::sync_with_stdio(0);
  int i, n, k;
  cin >> n >> k;
  for (i = 1; i <= n; ++i) cin >> a[i];
  if (n == 1) {
    cout << a[1];
    return 0;
  }
  sort(a + 1, a + n + 1);
  i = n;
  int j = 0;
  int ct = 1;
  while (i != 0 && ct != k + 1) {
    b[++j] = a[i];
    --i;
    ++ct;
  }
  if (n <= k) {
    cout << b[1];
    return 0;
  } else {
    long long int ans = -1;
    int j = k + 1;
    while (i != 0) {
      b[--j] += a[i];
      --i;
    }
    for (i = 1; i <= k; ++i) ans = max(ans, b[i]);
    cout << ans;
  }
}
