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(int(1e6))

n, m = map(int, input().split())

edges = {}

for i in range(m):
    s, t = map(int, input().split())
    if s not in edges:
        edges[s] = set()
    if t not in edges:
        edges[t] = set()
    edges[s].add(t)
    edges[t].add(s)

order = [0] * n
lowlink = [0] * n
visited = set()
bridges = set()


def dfs(v, p, i):
    visited.add(v)

    i += 1
    lowlink[v] = order[v] = i

    if v not in edges:
        return

    for t in edges[v]:
        if t not in visited:
            dfs(t, v, i)
            lowlink[v] = min(lowlink[v], lowlink[t])
            if lowlink[t] == order[t]:
                bridges.add((min(v, t), max(v, t)))
        elif t != p:
            lowlink[v] = min(lowlink[v], lowlink[t])


k = 0
for i in range(n):
    if i not in visited:
        dfs(i, -1, k)

for s, t in sorted(bridges):
    print(s, t)