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 <bits/stdc++.h>
using namespace std;
#define dhoom ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);  
typedef long long ll;
#define SIZE 100005
vector< int > graph[SIZE];
vector< pair< int , int > > ans;
vector< int > vis(SIZE,0);
vector< int > low(SIZE,0);
vector< int > tin(SIZE , 0);

int timer;
void dfs(int u , int p = -1){
	vis[u] = true;
	tin[u] = low[u] = timer++;
	int children = 0;
	for(auto it : graph[u]){
			if(it == p) continue;
			if(vis[it] == 1)
				low[u] = min(low[u] , tin[it]);
			else{
				dfs(it , u);
				low[u] = min(low[u] , low[it]);
				if(low[it] > tin[u])
					it > u ? ans.push_back({u, it}) : ans.push_back({it,u});
				children++;
			}
	}
}
int main(){
	dhoom;
	int n , m;
	cin >> n >> m;
	for(int i = 0 ; i < m ; i++){
		int x , y ;
		cin >> x >> y;
		graph[x].push_back(y);
		graph[y].push_back(x);
	}
	for(int i = 0 ; i < n ; i++){
			if(!vis[i])
				dfs(i);
	}
	sort(ans.begin() , ans.end());
	ans.erase(unique(ans.begin() , ans.end()), ans.end());
	for(int i = 0 ; i < (int) ans.size() ; i++)	
		cout << ans[i].first<< " " << ans[i].second << endl;
    return 0;
}