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<map>
#include<set>

using namespace std;

vector<int> E[100010];
vector<int> res;
int o[100010];
vector<pair<int,int> > bridge;


int dfs(int i,int prev,int &n) {
    if(o[i]) return o[i];

    int M=n;
    o[i] = n++;
    for(vector<int>::iterator it = E[i].begin(); it != E[i].end(); it++) {
        if(prev == *it) continue;
        if(o[i] < o[*it]) continue;

        int res = dfs(*it, i, n);
        M = min(M, res);

        if(o[i] < res) bridge.push_back(pair<int,int>(min(i,*it),max(i,*it)));
    }
    return M;
}

int main(void) {
	int v,e;
	scanf("%d%d",&v,&e);
	for(int i=0; i<e; i++) {
		int s,t;
		scanf("%d%d",&s,&t);
		E[s].push_back(t);
		E[t].push_back(s);
	}

    int n=1;
	dfs(0,-1,n);

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

    for(vector<pair<int,int> >::iterator it = bridge.begin(); it != bridge.end(); it++) {
        printf("%d %d\n",it->first,it->second);
    }
    return 0;
}