Recently Duff has been a soldier in the army. Malek is her commander.
Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.
There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is iand he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.
Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.
To answer a query:
Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order.
If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).
Duff is very busy at the moment, so she asked you to help her and answer the queries.
The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).
The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).
Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).
Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).
For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.
5 4 5 1 3 1 2 1 4 4 5 2 1 4 3 4 5 6 1 5 2 5 5 10 2 3 3 5 3 1
1 3 2 2 3 0 3 1 2 4 1 2
Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):
题目大意：有n个城市，形成一棵树，有若干个人，按照输入顺序编号，给出它们所在的城市。有q个询问，对于每个询问$\left(u,v\right)$输出在城市u、v之间的最短路径上的所有城市中编号最小的a个人。
因为a最大只有10，所以可以在做倍增LCA的预处理时处理出路径上编号最小的10个人，在查询时进行合并就行了。
代码在此。
Nov 23, 2015 09:12:10 PM
艹。。然而我当时写的是dfs序主席树
Dec 04, 2015 09:57:02 PM
@qiancl: %