【31.72%】【codeforces 604B】More Cowbell

time limit per test2 seconds

memory limit per test256 megabytes

inputstandard input

outputstandard output

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}.

【题目链接】:http://codeforces.com/contest/604/problem/B

【题解】



二分最后的箱子容量;

左端点应该是最大的cowbell,右端点无限大.

看看m需要用几个箱子d;

(如果加上这个数大于m或装了两个就不装了);

(装的时候，先装大的,大的尝试和当前剩余最小的组合在一起,如果能组合就组合,不能的话大的单独装.);

if (d <= k)

ans = m,r = m-1;

else

if (d > k)//箱子用多了那就增大箱子容量

l = m+1;

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int MAXN = 1e5+100;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);

int n,k;
int s[MAXN];

int ok(int lim)
{
    int tot = 0;
    int r = n,l = 1;
    while (l <= r)
    {
        if (l!=r && s[r]+s[l]<=lim)
        {
            r--,l++;
            tot++;
        }
        else
            tot++,r--;
    }
    return tot;
}

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    rei(n);rei(k);
    int l = 1,r = 21e8;
    rep1(i,1,n)
        {
            rei(s[i]);
            l = max(s[i],l);
        }
    int ans = -1;
    while (l <= r)
    {
        int m = (l+r)>>1;
        if (ok(m)<=k)
        {
            ans = m;
            r = m-1;
        }
        else
            l = m+1;
    }
    cout << ans << endl;
    return 0;
}