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;
typedef long long int ll;
typedef vector<ll> vl;
typedef pair<ll,ll> P;
#define rep(i,n) for(ll i = 0; i < ll(n); i++)
#define all(v) v.begin() , v.end()
#define inputv(v,n) rep(i,n){\
    ll x;\
    cin>>x;\
    v.push_back(x);\
    }
const ll INF=99999999999999;
const ll MOD=1000000007;
ll c,d,e,f,p,t,x,y,ans=0;
 
set<ll> s,loop;
vl visited(300000,0);
ll n,m,r;
ll num=0;
vl gragh[300000];
vl low(300000,INF),pre(300000,INF);
vector<P> ANS;
 
void input(void){
    cin>>n>>m;
    rep(i,m){
        cin>>c>>d;
        gragh[d].push_back(c);
        gragh[c].push_back(d);
    }
}

void dfs(ll a,ll b){
    visited[a]=1;
    pre[a]=num;
    low[a]=num;
    num++;
    for(ll i:gragh[a]){
        if(i!=b){
            if(visited[i]==0){
                dfs(i,a);
            }
            if(pre[a]<pre[i])low[a]=min(low[a],low[i]);
		else low[a]=min(low[a],pre[i]);
        }
    }

    if(pre[a]!=0&&pre[a]==low[a]){
        ANS.emplace_back(min(a,b),max(a,b));
    }
}

ll solve(void){
    dfs(0,-1);
    sort(all(ANS));
    rep(i,ANS.size()){
        x=ANS[i].first;
        y=ANS[i].second;
        cout<<x<<" "<<y<<endl;
    }
    return ans;
}
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    input();
    solve();
}