// Just GRL_3_A with added search for bridges
#include <bits/stdc++.h>
using namespace std;

typedef long long int ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<string> vs;
typedef vector<vb> vvb;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
#define all(x) x.begin(), x.end()
#define rep(i,a,b) for(int i = a; i < b; i++)

int V, E;
vvi conn(100002);
vi vis(100002);
vb trail(100002);
vpii bridges(0);

int dfs_min(int v, int depth, int parent) {
  if(vis[v] > 0) return vis[v];
  vis[v] = depth;
  trail[v] = true;

  int minHit = depth;
  rep(i,0,(int)conn[v].size()) {
    if(conn[v][i] != parent && (vis[conn[v][i]] == 0 || trail[conn[v][i]])) {
      int hit = dfs_min(conn[v][i], depth+1, v);
      minHit = min(minHit, hit);
      if(hit > depth)
	bridges.push_back(make_pair(min(v,conn[v][i]), max(v,conn[v][i])));
    }
  }

  trail[v] = false;
  return minHit;
}

int main() {
  cin >> V;
  cin >> E;
  rep(e,0,E) {
    int s, t;
    cin >> s;
    cin >> t;
    conn[s].push_back(t);
    conn[t].push_back(s);
  }

  dfs_min(0, 1, -1);
  /*if(V > 1) {
    cut[0] = false;
    bool remember = cut[1];
    fill(all(vis), 0);
    dfs_min(1, 1, -1);
    cut[1] = remember;
    }*/


  /*rep(v,0,V) {
    if(!(cut[v] || conn[v].size() == 1)) continue;
    rep(i,0,(int)conn[v].size())
      if(v < conn[v][i] && (cut[conn[v][i]] || conn[conn[v][i]].size() == 1))
	bridges.push_back(to_string(v)+" "+to_string(conn[v][i]));
	}*/
  sort(all(bridges));
  rep(b,0,(int)bridges.size())
    cout << bridges[b].first << ' ' << bridges[b].second << '\n';
}


### Prompt

Please formulate a CPP solution to the following 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

### Response

```cpp
// Just GRL_3_A with added search for bridges
#include <bits/stdc++.h>
using namespace std;

typedef long long int ll;
typedef pair<int,int> pii;
typedef pair<ll,ll> pll;
typedef vector<bool> vb;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef vector<pii> vpii;
typedef vector<pll> vpll;
typedef vector<string> vs;
typedef vector<vb> vvb;
typedef vector<vi> vvi;
typedef vector<vll> vvll;
#define all(x) x.begin(), x.end()
#define rep(i,a,b) for(int i = a; i < b; i++)

int V, E;
vvi conn(100002);
vi vis(100002);
vb trail(100002);
vpii bridges(0);

int dfs_min(int v, int depth, int parent) {
  if(vis[v] > 0) return vis[v];
  vis[v] = depth;
  trail[v] = true;

  int minHit = depth;
  rep(i,0,(int)conn[v].size()) {
    if(conn[v][i] != parent && (vis[conn[v][i]] == 0 || trail[conn[v][i]])) {
      int hit = dfs_min(conn[v][i], depth+1, v);
      minHit = min(minHit, hit);
      if(hit > depth)
	bridges.push_back(make_pair(min(v,conn[v][i]), max(v,conn[v][i])));
    }
  }

  trail[v] = false;
  return minHit;
}

int main() {
  cin >> V;
  cin >> E;
  rep(e,0,E) {
    int s, t;
    cin >> s;
    cin >> t;
    conn[s].push_back(t);
    conn[t].push_back(s);
  }

  dfs_min(0, 1, -1);
  /*if(V > 1) {
    cut[0] = false;
    bool remember = cut[1];
    fill(all(vis), 0);
    dfs_min(1, 1, -1);
    cut[1] = remember;
    }*/


  /*rep(v,0,V) {
    if(!(cut[v] || conn[v].size() == 1)) continue;
    rep(i,0,(int)conn[v].size())
      if(v < conn[v][i] && (cut[conn[v][i]] || conn[conn[v][i]].size() == 1))
	bridges.push_back(to_string(v)+" "+to_string(conn[v][i]));
	}*/
  sort(all(bridges));
  rep(b,0,(int)bridges.size())
    cout << bridges[b].first << ' ' << bridges[b].second << '\n';
}


```