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:
// 以下のライブラリを使ってみる.
// dohatsu氏.
// http://judge.u-aizu.ac.jp/onlinejudge/review.jsp?rid=1927353#1
#include<bits/stdc++.h>
using namespace std;
typedef pair<int, int> P;
#define MAX 100005
int V, E;
vector<int> G[MAX];
int depth[MAX];
int cnt[MAX];
int visited[MAX];
vector<P> ans;

void dfs(int pos, int prev){
    visited[pos] = true;
    for(int i = 0; i < (int)G[pos].size(); i++){
        int to = G[pos][i];
        if(to==prev)continue;
        if(!visited[to]){
            depth[to] = depth[pos] + 1;
            dfs(to, pos);
            cnt[pos]+=cnt[to];
            if(cnt[to] == 0) ans.push_back(P(min(pos,to) , max(pos,to)));
        }else if(depth[to] < depth[pos]){
            cnt[pos]++;
            cnt[to]--;
        }
    }
}
 
int main(){
    scanf("%d %d", &V, &E);
    for(int i = 0; i < E; i++){
        int a, b;
        scanf("%d %d", &a, &b);
        G[a].push_back(b);
        G[b].push_back(a);
    }
    dfs(0,-1);
    sort(ans.begin(), ans.end());
    for(int i = 0; i < (int)ans.size(); i++) printf("%d %d\n",ans[i].first, ans[i].second);
    return 0;
}