codeforces604B

More Cowbell

 CodeForces - 604B 

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.

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}.

sol：很显然答案是可以二分的，难点在于判断当前答案是否可行，一种较为容易想到的贪心，尽量用一个最大的配上一个最小的，易知一定是最优的

#include <bits/stdc++.h>
using namespace std;
typedef int ll;
inline ll read()
{
    ll s=0;
    bool f=0;
    char ch=' ';
    while(!isdigit(ch))
    {
        f|=(ch=='-'); ch=getchar();
    }
    while(isdigit(ch))
    {
        s=(s<<3)+(s<<1)+(ch^48); ch=getchar();
    }
    return (f)?(-s):(s);
}
#define R(x) x=read()
inline void write(ll x)
{
    if(x<0)
    {
        putchar('-'); x=-x;
    }
    if(x<10)
    {
        putchar(x+'0'); return;
    }
    write(x/10);
    putchar((x%10)+'0');
    return;
}
#define W(x) write(x),putchar(' ')
#define Wl(x) write(x),putchar('
')
const int N=100005;
int n,m,a[N];
inline bool Judge(int mid)
{
    int l=1,r=n,cnt=0;
    while(l<=r)
    {
        if(a[l]+a[r]<=mid)
        {
            cnt++; l++; r--;
        }
        else
        {
            cnt++; r--;
        }
    }
    return (cnt<=m)?1:0;
}
int main()
{
    int i;
    R(n); R(m);
    for(i=1;i<=n;i++) R(a[i]);
    sort(a+1,a+n+1);
    int l=a[n],r=2000000;
    while(l<=r)
    {
        int mid=(l+r)>>1;
        if(Judge(mid)) r=mid-1;
        else l=mid+1;
    }
    Wl(l);
    return 0;
}
/*
input
2 1
2 5
output
7

input
4 3
2 3 5 9
output
9

input
3 2
3 5 7
output
8
*/