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 $1, 2, ... n$.

Constraints

* $1 \leq n \leq 100$

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 $u$.

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.

Example

Input

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


Output

1 0
2 1
3 2
4 1
import sys
n = int(input())
input_line = [[int(i) for i in str.split()] for str in sys.stdin.read().splitlines()]
for i in range(n):
    input_line[i] = [] if len(input_line[i]) == 2 else input_line[i][2:]
is_visit = [False] * n
distance = [0] * n
queue = [1]

while len(queue) != 0:
    id = queue.pop(0)
    is_visit[id-1] = True
    for node_id in input_line[id-1]:
        if not is_visit[node_id-1]:
            distance[node_id-1] = distance[id-1] + 1
            queue.append(node_id)
            is_visit[node_id-1] = True

for i in range(n):
    d = distance[i] if is_visit[i] else -1
    print('{0} {1}'.format(i+1, d))




