import sys
sys.stdin = open('data.txt', 'r')from collections import dequen, m = map(int, input().split())
arr = []
for i in range(n):arr.append(int(input()))'''
dp(i)表示前i头牛的合法选择中最大总价值的数值
用最后一头牛的状态切分状态
1. 最后一头牛不选 子集1的最大价值就是dp(i-1)
2. 最后一头牛选，以结尾连续的牛个数进行子集划分, 设S(i)是前缀和最大价值是max {dp(i-2) + S(i) - S(i-1),dp(i-3) + S(i) - S(i-2),......dp(i-m-1) + S(i) - S(i-m) }那么只需要求i位置前面长度为m的滑动窗口中，dp(i-2)-S(i-1)这个序列在窗口中的最大值即可
'''def solve(arr, n, m):if m <= 0:return 0S = [val for val in arr]for i in range(1, n):S[i] += S[i-1]que = deque()dp = [0] * ndp[0] = arr[0]for i in range(1, n):if i <= m-1:dp[i] = S[i]elif i == m:dp[i] = max(dp[i-1], S[i] - min(S[:m]))else:dp[i] = max(dp[i-1], que[0][1] + S[i])# 更新单调队列new_val = dp[i-1] - S[i]while len(que) > 0 and i - que[0][0] >= m:que.popleft()while len(que) > 0 and new_val >= que[-1][1]:que.pop()que.append( (i, new_val) )return dp[n-1]print(solve(arr, n, m))