'''
将三类物品转换成多重背包的物品即可，最后问题转换成
统一的多重背包问题，N^2复杂度求解
'''


from collections import deque

arr = []
N, V = map(int, input().split())
for _ in range(N):
    v, w, s = map(int, input().split())
    arr.append((v, w, s))

dp = [[0] * (V + 1) for _ in range(N)]

for i in range(N):
    # 开销，价值，最大数量
    ci, wi, type = arr[i]

    if type == -1:
        si = 1
    elif type == 0:
        si = V // ci
    else:
        si = type

    if i == 0:
        for j in range(V + 1):
            cnt = min(j // ci, si)
            dp[i][j] = cnt * wi
    else:
        for window_start in range(0, min(V, ci)):

            dp[i][window_start] = dp[i - 1][window_start]
            que = deque()
            que.append((dp[i - 1][window_start], 0))  # [入队数值，入队时间戳]
            time_cnt = 0

            for j in range(window_start + ci, V + 1, ci):
                time_cnt += 1

                # 删除过期数据
                while len(que) > 0 and time_cnt - que[0][1] >= si + 1:
                    que.popleft()

                while len(que) > 0 and que[-1][0] + (time_cnt - que[-1][1]) * wi <= dp[i - 1][j]:
                    que.pop()
                que.append((dp[i - 1][j], time_cnt))

                dp[i][j] = que[0][0] + (time_cnt - que[0][1]) * wi

print(dp[N - 1][V])