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 ll;
typedef unsigned long long ull;
typedef pair<ll, ll> P;

vector<vector<int> > g;
vector<int> ord, low;
vector<bool> is_articulation;
vector<P> bridges;

void dfs(int u, int par, int &k){
    ord[u] = k;
    k++;
    low[u] = ord[u];
    int cnt = 0;
    for(auto v : g[u]){
        if(ord[v] == -1){
            cnt++;
            dfs(v, u, k);
            low[u] = min(low[u], low[v]);
            if(par!=-1 && ord[u] <= low[v]){
                is_articulation[u] = true;
            }
            if(ord[u] < low[v]){
                bridges.push_back(make_pair(min(u,v), max(u,v)));
            }
        }
        else if(v != par)low[u] = min(low[u], ord[v]);
    }
    if(par == -1){
        is_articulation[u] = cnt >= 2;
    }
}

int main() {
    int n, m;
    cin >> n >> m;
    g.resize(n);
    ord.resize(n, -1);
    low.resize(n);
    is_articulation.resize(n, false);
    for(int i=0;i<m;i++){
        int a, b;
        cin >> a >> b;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    int k = 0;
    dfs(0, -1, k);
    sort(bridges.begin(), bridges.end());
    for(auto e : bridges){
        cout << e.first << " " << e.second << endl;
    }
    return 0;
}