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;

const int N=1e6+10;
vector<int> G[N];
int order[N];
int low[N];
int p[N];
int child=0;
int cnt=0;
vector<pair<int,int> > ans;

void add(int s,int t)
{
    G[s].push_back(t);
    G[t].push_back(s);
}

void dfs(int v)
{
    order[v]=low[v]=++cnt;
    for(int i=0;i<G[v].size();i++)
    {
        int u=G[v][i];
        if(!order[u])
        {
            if(v==0) child++;
            p[u]=v;
            dfs(u);
            low[v]=min(low[v],low[u]);
            if(low[u]>order[v]) ans.push_back(make_pair(min(u,v),max(u,v)));
        }else if(p[v]!=u)
        {
            low[v]=min(low[v],order[u]);
        }
    }
}

int main()
{
    int v,e;
    cin>>v>>e;

    for(int i=0;i<e;i++){
        int s,t;
        cin>>s>>t;
        add(s,t);
    }

    dfs(0);
    sort(ans.begin(),ans.end());
    for(int i=0;i<ans.size();i++)
    {
        cout<<ans[i].first<<" "<<ans[i].second<<endl;
    }

    return 0;
}