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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_iare 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.

Example

Input

2 12 5

Output

7

Input

4 32 3 5 9

Output

9

Input

3 23 5 7

Output

8

直接二分的。然后check的时候贪心的方法是尽量找两个物品总容量接近检查的mid值。

#include <bits/stdc++.h>using namespace std;const int MAXN=1e5+7;const int inf=1e9;int n,m,k;int a[MAXN];bool vis[MAXN];bool check(int mid){    int cnt=0;int p=n-1;    for(int i=0;i<n;++i)    {        if(vis[i])continue;        while((vis[p]||a[i]+a[p]>mid)&&p>=0)p--;        vis[i]=1;        if(p>=0)vis[p]=1;        cnt++;    }    if(cnt<=m)return 1;    else return 0;}int main(){    scanf("%d%d",&n,&m);    int low=0,high=0;    for(int i=0;i<n;++i)    {        scanf("%d",&a[i]);    }    low = a[n-1];    high=a[n-1]+a[n-2];    int ans;    while(low<=high)    {        int mid=(low+high)/2;        for(int i=0;i<n;++i)vis[i]=0;        if(check(mid))        {            ans=mid;            high=mid-1;        }        else low=mid+1;    }    printf("%d\n",ans);    return 0;}