'''
先用贪心进行排序，能量损失小的石头先选取，
在排序后的序列上进行01背包
'''

T = int(input())

from functools import total_ordering
@total_ordering
class stone:
    def __init__(self, s, e, l):
        self.s = s
        self.e = e
        self.l = l

    def __eq__(self, other):
        return (self.s, self.e, self.l) == (other.s, other.e, other.l)

    def __lt__(self, other):
        val1 = self.l * other.s
        val2 = other.l * self.s
        return val1 < val2

    def __repr__(self):
        return f'{self.s},{self.e},{self.l}'

for case_id in range(1, T+1):
    N = int(input())
    arr = []
    for i in range(N):
        s, e, l = map(int, input().split())
        arr.append( stone(s, e, l) )

    # 首先进行排序, 最后的答案一定是在排序之后的序列中进行01背包
    arr.sort()
    total_j = sum([stone.s for stone in arr])

    # dp(i, j) 表示在前i种石头中做选择，剩余时间为j的约束下所有选法中最大总价值的数值
    dp = [0] * (total_j + 1)
    for i in range(N):
        for j in range(total_j, -1, -1):
            left_e = max(arr[i].e - (total_j - j) * arr[i].l, 0)
            if i == 0:
                dp[j] = left_e if j >= arr[i].s else 0
            else:
                if j >= arr[i].s:
                    dp[j] = max(dp[j], dp[j-arr[i].s] + left_e)
    print(f'Case #{case_id}: {dp[total_j]}')