Problem: Find bridges of an undirected graph G(V, E).

A bridge (also known as a cut-edge) is an edge whose deletion increase the number of connected components.

Constraints

* 1 ≤ |V| ≤ 100,000
* 0 ≤ |E| ≤ 100,000
* The graph is connected
* There are no parallel edges
* There are no self-loops

Input


|V| |E|
s0 t0
s1 t1
:
s|E|-1 t|E|-1


, where |V| is the number of nodes and |E| is the number of edges in the graph. The graph nodes are named with the numbers 0, 1,..., |V|-1 respectively.

si and ti represent source and target nodes of i-th edge (undirected).

Output

A list of bridges of the graph ordered by name. For each bridge, names of its end-ponints, source and target (source < target), should be printed separated by a space. The sources should be printed ascending order, then the target should also be printed ascending order for the same source.

Examples

Input

4 4
0 1
0 2
1 2
2 3


Output

2 3


Input

5 4
0 1
1 2
2 3
3 4


Output

0 1
1 2
2 3
3 4

Solution (in PYTHON3):

Solution:
import sys

sys.setrecursionlimit(200000)

v, e = map(int, input().split())

edges = [set() for _ in range(v)]

for _ in range(e):
    s, t = map(int, input().split())
    edges[s].add(t)
    edges[t].add(s)

prenum = [None] * v
parent = [None] * v
lowest = [None] * v
counter = 0
bridges = set()

def dfs(cur, prev):
    global counter
    prenum[cur] = lowest[cur] = counter
    counter += 1
    for edge in edges[cur]:
        if prenum[edge] is not None:
            if edge != prev:
                lowest[cur] = min(lowest[cur], prenum[edge])
            continue
        parent[edge] = cur
        dfs(edge, cur)
        lowest[cur] = min(lowest[cur], lowest[edge])
    if prenum[cur] == lowest[cur]:
        bridges.add((cur, prev) if cur < prev else (prev, cur))
 
 
dfs(0, -1)
 
for bridge in sorted(bridges)[1:]:
    print(*bridge)