Ilya Muromets（DP or 思维）

Ilya Muromets

 Gym - 100513F

Силачом слыву недаром — семерых одним ударом!

From the Russian cartoon on the German fairy tale.

Ilya Muromets is a legendary bogatyr. Right now he is struggling against Zmej Gorynych, a dragon with n heads numbered from 1 to n from left to right.

Making one sweep of sword Ilya Muromets can cut at most k contiguous heads of Zmej Gorynych. Thereafter heads collapse getting rid of empty space between heads. So in a moment before the second sweep all the heads form a contiguous sequence again.

As we all know, dragons can breathe fire. And so does Zmej Gorynych. Each his head has a firepower. The firepower of the i-th head is f_i.

Ilya Muromets has time for at most two sword sweeps. The bogatyr wants to reduce dragon's firepower as much as possible. What is the maximum total firepower of heads which Ilya can cut with at most two sword sweeps?

Input

The first line contains a pair of integer numbers n and k (1 ≤ n, k ≤ 2·10⁵) — the number of Gorynych's heads and the maximum number of heads Ilya can cut with a single sword sweep. The second line contains the sequence of integer numbers f₁, f₂, ..., f_n (1 ≤ f_i ≤ 2000), where f_i is the firepower of the i-th head.

Output

Print the required maximum total head firepower that Ilya can cut.

Examples

Input

8 2
1 3 3 1 2 3 11 1

Output

20

Input

4 100
10 20 30 40

Output100

题意：
　　给你两个数n和k，然后n个数，进行两次操作，一次最多可以消除连续的K个数，然后剩余的数又变成连续的，问两次操作后，消除的数的总和最大为多少。
题解：
　　dp[i][1]表示的是1-i的区间中，第一次操作能得到的最大值，因为i从k-n,每次加一，所以每次对最大值可能造成改变的只有新加进来的数，dp[i][1]=max(dp[i-1][1],sum[i]-sum[i-k]);
　　dp[i][1]表示的是1-i的区间中，两次操作后能得到的最大值，dp[i][2]=max(dp[i-1][2],sum[i]-sum[i-k]+dp[i-k][1]);对于两次操作，状态转移方程就是1- i-1区间两次操作后最大值或者是新加入的i影响的区间sum[i]-sum[i-k]+1- i-k区间操作一次的最大值。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=2e5+10;
int sum[maxn],a[maxn],dp[maxn][3];
int main()
{
    int n,k;
    scanf("%d%d",&n,&k);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        sum[i]=sum[i-1]+a[i];
    }
    if(2*k>=n)
    {
        printf("%d\n",sum[n]);
        return 0;
    }
    memset(dp,0,sizeof(dp));
    for(int i=k;i<=n;i++)
    {
        dp[i][1]=max(dp[i-1][1],sum[i]-sum[i-k]);
        dp[i][2]=max(dp[i-1][2],sum[i]-sum[i-k]+dp[i-k][1]);
       // printf("%d %d %d\n",i,dp[i][1],dp[i][2]);
    }
    printf("%d\n",dp[n][2]);
    return 0;
}