'''
朴素思路dp(i, j) 表示在前i个分组中选择，总容量为j的条件下的最大价值和
枚举每个分组中选择哪个物品的状态即可'''N, V = map(int, input().split())arr = []
for i in range(N):arr.append([])Si = int(input())for j in range(Si):v, w = map(int, input().split())arr[i].append((v, w))dp = [[0 for _ in range(110)] for _ in range(110)]
for i in range(N):for j in range(V + 1):if i == 0:for k in range(len(arr[i])):if j >= arr[i][k][0]:dp[i][j] = max(dp[i][j], arr[i][k][1])else:dp[i][j] = dp[i - 1][j]for k in range(len(arr[i])):if j >= arr[i][k][0]:dp[i][j] = max(dp[i][j], dp[i - 1][j - arr[i][k][0]] + arr[i][k][1])print(dp[N - 1][V])