python

import heapq

n, m = map(int, input().split())
w = list(map(int, input().split()))

# 小根堆，维护每个水龙头的结束时间
heap = []

# 前 m 个同学直接占用所有水龙头
for i in range(min(m, n)):
    heapq.heappush(heap, w[i])

# 后续同学：取出最早结束的水龙头，加上该同学的接水时间
for i in range(m, n):
    earliest = heapq.heappop(heap)
    heapq.heappush(heap, earliest + w[i])

# 总时间 = 水龙头中的最大结束时间
print(max(heap) if heap else 0)