一般贪心

Description

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.

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, ..., 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.

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.

Sample Input

Input

2 12 5

Output

7

Input

4 32 3 5 9

Output

9

Input

3 23 5 7

Output

8

Sample Output

Hint

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}.

题意大概是N个东西放到K个箱子，可以两个放到一起，求最小的箱子体积（没个箱子都一样大）

此为一道贪心，然而直接思考可能太抽象，不妨把想象过程，在思想中模拟（神探夏洛克的的思维宫殿),这时就会发现只是一道水题#.#;

我们先从大的开始放到箱子里，直到放满，剩下的小的再往里放，最小的放到能相对来说最大的，

贪心就是找到一种放的方式先开始，然后开始规划改变称最优的；

这里给一下代码

#include<stdio.h>
#include<algorithm>
using namespace std;
long long n,k,s[100005],maxs;
int main()
{
    scanf("%lld%lld",&n,&k);
    for(long long i=0; i<n; i++)
        scanf("%lld",&s[i]);
    maxs=s[n-1];
    for(long long i=0; i<n-k; i++)
        maxs=max(maxs,s[i]+s[(n-k)*2-i-1]);

    printf("%lld\n",maxs);
    return 0;
}