LCA - Lowest Common Ancestor

A tree is an undirected graph in which any two vertices are connected by exactly one simple path. In other words, any connected graph without cycles is a tree. - Wikipedia 

The lowest common ancestor (LCA) is a concept in graph theory and computer science. Let T be a rooted tree with N nodes. The lowest common ancestor is defined between two nodes v and w as the lowest node in T that has both v and w as descendants (where we allow a node to be a descendant of itself). - Wikipedia

Your task in this problem is to find the LCA of any two given nodes v and w in a given tree T.

For example the LCA of nodes 9 and 12 in this tree is the node number 3.

Input

The first line of input will be the number of test cases. Each test case will start with a number N the number of nodes in the tree, 1 <= N <= 1,000. Nodes are numbered from 1 to N. The next N lines each one will start with a number M the number of child nodes of the Nth node, 0 <= M <= 999 followed by M numbers the child nodes of the Nth node. The next line will be a number Q the number of queries you have to answer for the given tree T, 1 <= Q <= 1000. The next Q lines each one will have two number v and w in which you have to find the LCA of v and w in T, 1 <= v, w <= 1,000.

Input will guarantee that there is only one root and no cycles.

Output

For each test case print Q + 1 lines, The first line will have “Case C:” without quotes where C is the case number starting with 1. The next Q lines should be the LCA of the given v and w respectively.

Example

Input:
1
7
3 2 3 4
0
3 5 6 7
0
0
0
0
2
5 7
2 7

Output:
Case 1:
3
1

Được gửi lên bởi:hossamyosef
Ngày:2013-05-13
Thời gian chạy:0.600s-1.113s
Giới hạn mã nguồn:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Ngôn ngữ cho phép:Tất cả ngoại trừ: GOSU PERL6 PYPY RUST SED
Nguồn bài:FCIS/ASU Local Contest 2013

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2019-12-24 14:53:00
SIGSEGV =.=
2019-01-10 04:15:58
Nhật Hào hết sạch r à :vvvv
2018-07-23 03:42:15
code acc
https://ideone.com/z4aEBL
2018-07-23 03:38:14
Nhật Hào dơ
2017-12-30 10:33:24
ledacthuongvq
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