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;
static const double EPS = 1e-9;
template <class T>
bool INRANGE(T x, T a, T b) {
  return a <= x && x <= b;
}
template <class T>
void amin(T &a, T v) {
  if (a > v) a = v;
}
template <class T>
void amax(T &a, T v) {
  if (a < v) a = v;
}
int ROUND(double x) { return (int)(x + 0.5); }
bool ISINT(double x) { return fabs(ROUND(x) - x) <= EPS; }
bool ISEQUAL(double x, double y) {
  return fabs(x - y) <= EPS * max(1.0, max(fabs(x), fabs(y)));
}
double SQSUM(double x, double y) { return x * x + y * y; }
int main() {
  int n, k;
  scanf("%d %d ", &n, &k);
  vector<int> s(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d ", &s[i]);
  }
  long long lo = 0;
  long long hi = 1e10;
  {
    while (lo < hi) {
      long long mid = lo + (hi - lo) / 2LL;
      vector<int> pack0(k);
      vector<int> pack1(k, (-1));
      int id0 = 0;
      int id1 = k - 1;
      bool ok = true;
      if (s[n - 1] > mid) {
        ok = false;
      } else {
        for (int i = n - 1; i >= 0; --i) {
          if (id0 < k && s[i] <= mid) {
            pack0[id0] = s[i];
            id0++;
          } else {
            bool ireta = false;
            while (id1 >= 0) {
              if (pack1[id1] == (-1) && pack0[id1] != (-1) &&
                  pack0[id1] + s[i] <= mid) {
                pack1[id1] = s[i];
                ireta = true;
                break;
              }
              id1--;
            }
            if (!ireta) {
              ok = false;
              break;
            }
          }
        }
      }
      if (ok) {
        hi = mid;
      } else {
        lo = mid + 1;
      }
    }
  }
  cout << lo << endl;
  return 0;
}
