Problem:
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Breadth First Search

Write a program which reads an directed graph $G = (V, E)$, and finds the shortest distance from vertex $1$ to each vertex (the number of edges in the shortest path). Vertices are identified by IDs $1, 2, ... n$.

Input

In the first line, an integer $n$ denoting the number of vertices, is given. In the next $n$ lines, adjacent lists of vertex $u$ are given in the following format:

$u$ $k$ $v_1$ $v_2$ ... $v_k$

$u$ is ID of the vertex and $k$ denotes its degree.$v_i$ are IDs of vertices adjacent to $u$.

Constraints

$1 \leq n \leq 100$

Output

For each vertex $u$, print $id$ and $d$ in a line. $id$ is ID of vertex $u$ and $d$ is the distance from vertex $1$ to vertex $u$. If there are no path from vertex $1$ to vertex $u$, print -1 as the shortest distance. Print in order of IDs.

Sample Input 1

4
1 2 2 4
2 1 4
3 0
4 1 3

Sample Output 1

1 0
2 1
3 2
4 1

Reference

Introduction to Algorithms, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, and Clifford Stein. The MIT Press.

Solution in Ruby (judge verdict: Accepted)

INFTY = (1<<21)



def bfs(s)
	q = [s]
	d = Array.new($n, INFTY)
	d[s] = 0

	while !q.empty?
		u = q.shift
		$n.times do |v|
			next if $m[u][v] == 0
			next if d[v] != INFTY
			d[v] = d[u] + 1
			q.push(v)
		end
	end

	$n.times do |i|
		print i+1, " ", (d[i] == INFTY ? -1 : d[i]), "\n"
	end
end



$n = gets.to_i
$m = Array.new($n){ Array.new($n, 0) }

$n.times do
	u, k, *adj_list = gets.split.map &:to_i
	u -= 1
	adj_list.each do |v|
		v -= 1
		$m[u][v] = 1
	end
end

bfs(0)