Problem:
Score : 400 points

Problem Statement

There are N squares aligned in a row.
The i-th square from the left contains an integer a_i.

Initially, all the squares are white.
Snuke will perform the following operation some number of times:

Select K consecutive squares. Then, paint all of them white, or paint all of them black. Here, the colors of the squares are overwritten.

After Snuke finishes performing the operation, the score will be calculated as the sum of the integers contained in the black squares.
Find the maximum possible score.

Constraints

1≤N≤10^5

1≤K≤N

a_i is an integer.

|a_i|≤10^9

Input

The input is given from Standard Input in the following format:

N K
a_1 a_2 ... a_N

Output

Print the maximum possible score.

Sample Input 1

5 3
-10 10 -10 10 -10

Sample Output 1

10

Paint the following squares black: the second, third and fourth squares from the left.

Sample Input 2

4 2
10 -10 -10 10

Sample Output 2

20

One possible way to obtain the maximum score is as follows:

Paint the following squares black: the first and second squares from the left.

Paint the following squares black: the third and fourth squares from the left.

Paint the following squares white: the second and third squares from the left.

Sample Input 3

1 1
-10

Sample Output 3

0

Sample Input 4

10 5
5 -4 -5 -8 -4 7 2 -4 0 7

Sample Output 4

17

Solution in C (GCC 5.4.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
int main (void){
    int loop1,loop2;
    int how_data;
    int can;
    scanf("%d %d\n",&how_data,&can);
    long long int data[how_data];
    int data_flag[how_data]; //0を許さない
    //int data_flag2[how_data]; //0を許す
    long long int max=0;
    for(loop1=0;loop1<how_data;loop1++){
        scanf("%lld",&data[loop1]);
        if(data[loop1]>0)data_flag[loop1]=1; else data_flag[loop1]=0;
        //if(data[loop1]>=0)data_flag2[loop1]=1; else data_flag2[loop1]=0;
        if(data[loop1]>=0)max+=data[loop1];
    }
    //printf("%lld\n",max);
    //何回も塗れる→最終的に同一色がcan個並んでるのが1か所あればいい
    //なければどうする？
    int ren=0;
    int mae=2;
    int end=0;
    for(loop1=0;loop1<how_data;loop1++){
        if(mae==data_flag[loop1] || data[loop1]==0){
            ren++;
            if(ren==can){
                end=1;
            }
        }else ren=0;
        mae=data_flag[loop1];
    }
    mae=0;ren=0;
    /*for(loop1=0;loop1<how_data;loop1++){
        if(mae==data_flag2[loop1]){
            ren++;
            if(ren==can){
                end=1;
            }
        }else ren=0;
        mae=data_flag[loop1];
    }*/
    
    //printf("%d\n",end);
    
    //もし、まだend=0ならどうするか？
    //最も小さい正を白にする or 最も大きい負を黒にする
    //それだけでren==canになる保証はない
    //現在maxに格納されてる値は正の総和
    //canのビット数を全部0or1でしたときのmaxの変化を見る？
    
    long long int s_max;
    long long int k_max;
    long long int max2=-1000000000000000;
    if(end==0){
        for(loop1=0;loop1<how_data-can+1;loop1++){
            s_max=max;
            k_max=max;
            for(loop2=loop1;loop2<loop1+can;loop2++){
                if(data[loop2]>=0)s_max-=data[loop2];
                if(data[loop2]<=0)k_max+=data[loop2];
            }
            //全部黒に
            /*k_max=s_max;
            for(loop2=loop1;loop2<loop1+can;loop2++){
                k_max+=data[loop2];
            }*/
            //printf("%d～%d\n",loop1,loop1+can-1);
            //printf("k=%lld s=%lld\n",k_max,s_max);
            if(k_max>max2)max2=k_max;
            if(s_max>max2)max2=s_max;
        }
    }
    if(end==1)printf("%lld\n",max);else printf("%lld\n",max2);
    return 0;
}