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 Python (judge verdict: Wrong Answer)

n = int(raw_input())

d = [99999 for i in range(n)]
G = [0 for i in range(n)]
M = [[0 for i in range(n)] for j in range(n)]
g = [0 for i in range(n)]
frag = [[0 for i in range(n)] for j in range(n)]
temp_d = [0]

def BFS_visit(s, temp_d):
    if temp_d == 99999:
        temp_d = 0
    if d[s] > temp_d:
        d[s] = temp_d
    temp_d += 1
    for e in range(1, n):
        if M[s][e] == 1 and s != e and frag[s][e] == 0:
            frag[s][e] = 1
            BFS_visit(e, temp_d)

def BFS():
    s = 0
    BFS_visit(0, d[s])
    for s in range(n):
        if d[s] == 99999:
            d[s] = -1
    for s in range(n):
        print'{0} {1}'.format(s+1, d[s])

for i in range(n):
    G = map(int, raw_input().split())
    for j in range(G[1]):
        M[G[0]-1][G[2+j]-1] = 1
    g[i] = G[1]
BFS()