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;

#define MAX_V 100010
#define INF 1e9
typedef vector<int> Vec;
typedef pair<int,int> pii;

int low[MAX_V],ord[MAX_V];
int V,E;
Vec G[MAX_V];
vector<pii> bridge;

void dfs(int v,int prev,int &k){
    ord[v] = low[v] = k++;
    for(int i = 0 ; i < (int)G[v].size() ; i++){
        int to = G[v][i];
        if(to == prev) continue;
        if(ord[to] == -1){
            dfs(to, v, k);
            low[v] = min(low[v], low[to]);
            if(ord[v] < low[to]){
                int a = min(v, to),b = max(v, to);
                bridge.push_back(pii(a,b));
            }
        }
        low[v] = min(low[v], ord[to]);
    }
}

void init(){
    bridge.clear();
    for(int i = 0 ; i < V ; i++){
        G[i].clear();
        ord[i] = -1; low[i] = INF;
    }
}

int main(){
    int s,t;
    cin >> V >> E; init();
    for(int i = 0 ; i < E ; i++){
        cin >> s >> t;
        G[s].push_back(t);
        G[t].push_back(s);
    }
    int k = 0;
    for(int i = 0 ; i < V ; i++){
        if(ord[i] == -1){
            dfs(i, -1, k);
        }
    }
    sort(bridge.begin(),bridge.end());
    for(int i = 0 ; i < (int)bridge.size() ; i++){
        cout << bridge[i].first << " " << bridge[i].second << endl;
    }
    return 0;
}