Cities Excursions
Contest: Codeforces Round 436 (Div. 2)
Difficulty rating: 2700
Tags: dfs and similar, graphs, trees

There are n cities in Berland. Some pairs of them are connected with m directed roads. One can use only these roads to move from one city to another. There are no roads that connect a city to itself. For each pair of cities (x, y) there is at most one road from x to y.

A path from city s to city t is a sequence of cities p1, p2, ... , pk, where p1 = s, pk = t, and there is a road from city pi to city pi + 1 for each i from 1 to k - 1. The path can pass multiple times through each city except t. It can't pass through t more than once.

A path p from s to t is ideal if it is the lexicographically minimal such path. In other words, p is ideal path from s to t if for any other path q from s to t pi < qi, where i is the minimum integer such that pi ≠ qi.

There is a tourist agency in the country that offers q unusual excursions: the j-th excursion starts at city sj and ends in city tj.

For each pair sj, tj help the agency to study the ideal path from sj to tj. Note that it is possible that there is no ideal path from sj to tj. This is possible due to two reasons:

- there is no path from sj to tj;
- there are paths from sj to tj, but for every such path p there is another path q from sj to tj, such that pi > qi, where i is the minimum integer for which pi ≠ qi.

The agency would like to know for the ideal path from sj to tj the kj-th city in that path (on the way from sj to tj).

For each triple sj, tj, kj (1 ≤ j ≤ q) find if there is an ideal path from sj to tj and print the kj-th city in that path, if there is any.

Input format:
The first line contains three integers n, m and q (2 ≤ n ≤ 3000,0 ≤ m ≤ 3000, 1 ≤ q ≤ 4·105) — the number of cities, the number of roads and the number of excursions.

Each of the next m lines contains two integers xi and yi (1 ≤ xi, yi ≤ n, xi ≠ yi), denoting that the i-th road goes from city xi to city yi. All roads are one-directional. There can't be more than one road in each direction between two cities.

Each of the next q lines contains three integers sj, tj and kj (1 ≤ sj, tj ≤ n, sj ≠ tj, 1 ≤ kj ≤ 3000).

Output format:
In the j-th line print the city that is the kj-th in the ideal path from sj to tj. If there is no ideal path from sj to tj, or the integer kj is greater than the length of this path, print the string '-1' (without quotes) in the j-th line.

Constraints:
Time limit: 2.0 s
Memory limit: 256.0 MB

Example input:
7 7 5
1 2
2 3
1 3
3 4
4 5
5 3
4 6
1 4 2
2 6 1
1 7 3
1 3 2
1 3 5

Example output:
2
-1
-1
2
-1

Editorial:
There is a direct graph. For each query, you need to find the kj-th vertex in the lexicographically minimal path from sj to tj.
First group the queries on the vertex tj and find all vertices from which the vertex tj is achievable. For this you can invert all the arcs and run dfs from the vertex tj.
Now consider the query (sj, tj). For this query, you need to find the lexicographically minimal path from sj to tj. If the vertex tj is not achievable from sj, then the answer is '-1'. Otherwise, in the lexicographically minimal path p from sj to tj the vertex pi (i > 1) is the minimal vertex from vertices u such that there exists an arc (pi - 1, u) and tj is achievable from u.
Thus, we can build a new graph consisting of arcs satisfying the previous condition. Let us invert the arcs in this graph. Consider the vertices achievable from tj in this graph. They form an outgoing tree. Only for these vertices there is a lexicographically minimal path to tj. The lexicographically minimal path from the vertex sj to the vertex tj is equal to the inverted path from tj to sj in this tree.
So, we can use binary climb to get the k-th vertex on this path.