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 CPP):

Solution:
#include<cstdio>
#include<vector>
#include<algorithm>
#include<utility>
#include<numeric>
#include<iostream>
#include<array>
#include<string>
#include<sstream>
#include<stack>
#include<queue>
#include<list>
#include<functional>
#define _USE_MATH_DEFINES

#include<math.h>
#include<map>

#define SENTINEL 1000000001

#define min(a,b) ((a)>(b)?(b):(a))
#define max(a,b) ((a)>(b)?(a):(b))

#define INF 200000000

using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;

vector<int> G[100000];
vector<pii> bridges;
int ord[100000], low[100000];
bool checked[100000];

int k;

void dfs(int v, int p)
{
	checked[v] = true;
	ord[v] = k++;
	low[v] = ord[v];

	for (int i = 0; i != G[v].size(); i++)
	{
		if (!checked[G[v][i]])
		{
			dfs(G[v][i], v);
			low[v] = min(low[v], low[G[v][i]]);
			if (ord[v] < low[G[v][i]])
			{
				bridges.push_back(pii(min(v, G[v][i]), max(v, G[v][i])));
			}
		}
		else if (G[v][i] != p)
		{
			low[v] = min(low[v], ord[G[v][i]]);
		}
	}
}

int main() 
{
	cin.tie(0);
	ios::sync_with_stdio(false);

	int V, E;
	scanf("%d %d", &V, &E);

	for (int i = 0; i < E; i++)
	{
		int s, t;
		scanf("%d %d", &s, &t);

		G[s].push_back(t);
		G[t].push_back(s);
	}

	dfs(0, -1);

	sort(bridges.begin(), bridges.end());

	for (auto& p : bridges)
	{
		printf("%d %d\n", p.first, p.second);
	}
	return 0;
}