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 s_i. In fact, he keeps his cowbells sorted by size, so s_i - 1 ≤ s_i 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.

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, ..., s_n (1 ≤ s₁ ≤ s₂ ≤ ... ≤ s_n ≤ 1 000 000), the sizes of Kevin's cowbells. It is guaranteed that the sizes s_i are given in non-decreasing order.

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.

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}.

AC代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cstring>
using namespace std;
//#define LOCAL
#define MAX_N 100000int BOX[MAX_N];
int main()
{#ifdef LOCALfreopen("b:\\data.in.txt", "r", stdin);#endifint n , k;cin >> n >> k;for(int i = 0; i < n; i++){cin >> BOX[i] ;}//先给ans最大的量int ans = BOX[ n - 1 ];if( n <= k)cout << ans << endl ;else{//n > k && n <= 2*k时int len = n - k ;//记录前向遍历的长度int j = len ;for(int i = len-1 ; i >= 0 ; i--){ans = max( ans , BOX[i] + BOX[j++] );}cout << ans << endl ;}return 0;
}