0, 1, 2…i - j - 1, |i - j|, i - j + 1,....i;
j表示连续选择的长度j <= m,从i - j + 1 ~ i共连续选了j个数，则不选|i - j|

f(i):
不选i: f(i) = f(i - 1);
选择i: f(i) = f(i - j - 1) + s(i) - s(i - j);
|i-j| 右边根据前缀和公式s(r) - s(l - 1) 中: r = i, l = i - j + 1
|i-j| 左边即为 f(i - j - 1) 的最大值

选i时 f(i) = max(f(i - j - 1) + s(i) - s(i - j)) = max(f(i - j - 1) - s(i - j)) + s(i);
f(i - j - 1) - s(i - j) 使用 g(i - j)记录, j 表示长度, q[hh] = i - j 表示！起点前一个点位置！即g(q[hh]);

import java.util.*;
class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        long[] s = new long[n+1];
        for(int i = 1; i <= n; i++){
            s[i] = sc.nextInt();
            s[i] += s[i-1];
        }
        long[] f = new long[n+1];
        long[] g = new long[n+1];
        int[] q = new int[n+1];
        int hh = 0, tt = -1;            //将g[0]加入到队列中，否则会从g[1]开始考虑忽略g[0]
        q[++tt] = (int) g[0];
        for(int i = 1; i <= n; i++){
            if(hh <= tt && i - q[hh] > m) hh++;
            g[i] = f[i-1] - s[i];
            f[i] = Math.max(f[i-1], g[q[hh]] + s[i]);
            while(hh <= tt && g[q[tt]] <= g[i]) tt--;
            q[++tt] = i;
        }
        System.out.println(f[n]);
    }
}