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>
using namespace std;
struct edge{int to, index;};
static const int MAX_V = 100000;
static const int MAX_E = 100000;

int V, E;
vector<edge> G[MAX_V];
bool visited[MAX_V], passed[MAX_E];
int ord[MAX_V], low[MAX_V];

void dfs(int u, int k){
	visited[u] = true;
	ord[u] = k;
	k++;
	low[u] = ord[u];
	for(int i = 0; i < G[u].size(); i++){
		if(!visited[G[u][i].to]){
			passed[G[u][i].index] = true;
			dfs(G[u][i].to, k);
			low[u] = min(low[u], low[G[u][i].to]);
		}else if(!passed[G[u][i].index]){
			low[u] = min(low[u], ord[G[u][i].to]);
		}
	}
}

int main(){
	scanf("%d %d", &V, &E);
	for(int i = 0; i < E; i++){
		int s, t;
		scanf("%d %d", &s, &t);
		G[s].push_back((edge){t, i});
		G[t].push_back((edge){s, i});
	}
	fill(visited, visited + V, false);
	fill(passed, passed + E, false);
	dfs(0, 0);
	vector<pair<int, int> > res;
	for(int i = 0; i < V; i++){
		for(int j = 0; j < G[i].size(); j++){
			if(!passed[G[i][j].index]) continue;
			if(ord[G[i][j].to] < ord[i]) continue;
			if(ord[i] < low[G[i][j].to]){
				res.push_back(make_pair(min(i, G[i][j].to), max(i, G[i][j].to)));
			}
		}
	}
	sort(res.begin(), res.end());
	for(int i = 0; i < res.size(); i++) printf("%d %d\n", res[i].first, res[i].second);
	return 0;
}