A young farmer has N cows, but they produced really really a very very small amount of milk. John cannot live on the milk they made, so he’s planning to eat some of the ‘worst’ cows to get rid of hunger. Each day, John chooses the cow that produces the LEAST amount of milk on that day and eat them. If there are more than one cow with minimal milk, John will be puzzled and will not eat any of them (Yeah! That’s GREAT!!). The i-th cow has a cycle of production Ti. That means, if it produces L unit milk on one day, it will also produce L unit after Ti days — If it will not be eaten during these day :-). Though John is not a clever man, he doubts whether the cows will be eventually eaten up, so he asks for your help. Don’t forget that he will offer you some nice beef for that!
The first line of the input contains a single integer T , indicating the number of test cases (1 ≤ T ≤ 50). Each test case begins with an integer N (1 ≤ N ≤ 1000), the number of cows. In the following N lines, each line contains an integer Ti (1 ≤ Ti ≤ 10), indicating the cycle of the i-th cow, then Ti integers Mj (0 ≤ Mj ≤ 250) follow, indicating the amount of milk it can produce on the j-th day.
For each test case in the input, print a single line containing two integers C, D, indicating the number of cows that will NOT be eaten, and the number of days passed when the last cow is eaten. If no cow is eaten, the second number should be 0.
範例輸入 1
1 4 4 7 1 2 9 1 2 2 7 1 1 2
範例輸出 1
2 6
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