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 solve(vector<int> s, int k, int boxSize) {
  int pairs = s.size() - k;
  int singles = k - pairs;
  int size = s.size();
  if (pairs <= 0) {
    if (s[size - 1] < boxSize) {
      return -1;
    } else if (s[size - 1] == boxSize) {
      return 0;
    } else if (s[size - 1] > boxSize) {
      return 1;
    }
  } else if (pairs > 0) {
    vector<int> pairedOne;
    int i = 0;
    int j = pairs * 2 - 1;
    while (j > i) {
      int sum = s[i] + s[j];
      pairedOne.push_back(sum);
      i++;
      j--;
    }
    int t = 2 * pairs;
    while (t < s.size()) {
      pairedOne.push_back(s[t]);
      t++;
    }
    int max = INT_MIN;
    for (int k = 0; k < pairedOne.size(); k++) {
      if (max < pairedOne[k]) max = pairedOne[k];
    }
    if (max < boxSize) {
      return -1;
    } else if (max == boxSize) {
      return 0;
    } else if (max > boxSize) {
      return 1;
    }
  }
}
int binarySearch(vector<int> s, int k, int start, int end) {
  int ans = (start + end) / 2;
  int result = solve(s, k, ans);
  if (result == 1) {
    return binarySearch(s, k, ans + 1, end);
  }
  if (result == 0) {
    return ans;
  }
  if (result == -1) {
    return binarySearch(s, k, start, ans - 1);
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> s;
  for (int i = 0; i < n; i++) {
    int a;
    cin >> a;
    s.push_back(a);
  }
  if (n == 1) {
    cout << s[0];
  } else {
    int max = s[n - 1] + s[n - 2];
    cout << binarySearch(s, k, 1, max);
  }
}
