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:
//TO CHANGE THE WORLD
#include <bits/stdc++.h>

using namespace std;

#define pb push_back
#define mp make_pair
#define L first
#define R second

const int maxn=1e5+10;

vector <int> adj[maxn];
vector <pair<int, int> > ed;
bool mark[maxn];
int h[maxn], up[maxn], par[maxn];

void dfs(int x)
{
    up[x]=h[x];
    mark[x]=true;
    for(int i=0; i<adj[x].size(); i++)
    {
        int child=adj[x][i];
        if(!mark[child])
        {
            par[child]=x;
            h[child]=h[x]+1;
            dfs(child);
            up[x]=min(up[x],up[child]);
        }
        else if(child!=par[x])
            up[x]=min(up[x],h[child]);
    }
}

int main()
{
    int n,m;
    vector <pair <int, int> > ans;
    cin>>n>>m;
    for(int i=0; i<m; i++)
    {
        int fi,se;
        cin>>fi>>se;
        adj[fi].pb(se);
        adj[se].pb(fi);
        if(fi>se)
            swap(fi,se);
        ed.pb(mp(fi,se));
    }
    dfs(0);
    for(int i=0; i<m; i++)
    {
        int fi=ed[i].L, se=ed[i].R;
        if(h[fi]>h[se])
            swap(fi,se);
        if(up[se]>=h[se])
            ans.pb(mp(min(fi,se), max(fi,se)));
    }
    sort(ans.begin(), ans.end());
    for(int i=0; i<ans.size(); i++)
        cout<<ans[i].L<<" "<<ans[i].R<<endl;
    return 0;
}