#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

struct Edge {
    int s;
    int t;
    bool operator < (const Edge& tgt) const {
	if( s == tgt.s )
	    return t < tgt.t;
	return s < tgt.s;
    }
};



int Bridges_dfs(const vector<vector<Edge>>& g, vector<int>& order,
		vector<Edge>& ans, int u, int& cnt_v, Edge pre) {
    if( order[u] != -1 ) { //?¨????????????¨?????????
	return order[u];
    }
    // ??????????¨???????
    order[u] = cnt_v; cnt_v++;
    // ??£??\???????????????
    int min_v = order[u];
    for(auto e : g[u]) { // u ?????????????????????????????????????????????
	if( e.t == pre.s and e.s == pre.t ) continue; 
	min_v = min(Bridges_dfs(g, order, ans, e.t, cnt_v, e), min_v);
    }
    if( min_v == order[u] and pre.s != -1 ) {
	if(pre.s > pre.t) pre = Edge{pre.t,pre.s};
	ans.push_back(pre);
    }
    return min_v;
}

vector<Edge> Bridges(const vector<vector<Edge>>& g) {
    vector<Edge> ans;
    const int V = g.size();
    vector<int> order(V,-1);
    
    // 0??????DFS
    int cnt_v = 0;
    Edge tmp{-1,-1};
    Bridges_dfs(g, order, ans, 0, cnt_v, tmp);
    assert(cnt_v == V); // ??£???????????°?????????????????????
    return ans;
}

int main () {
    int V,E;
    cin >> V >> E;
    vector<vector<Edge>> g(V);
    for (int i=0; i<E; ++i) {
	int s,t;
	cin >> s >> t; 
	g[s].push_back(Edge{s,t});
	g[t].push_back(Edge{t,s});
    }
    vector<Edge> ans = Bridges(g);
    sort(ans.begin(), ans.end());
    for(auto& e : ans) {
	cout << e.s << " "<< e.t << endl;
    }

    return 0;
}
### Prompt

Generate 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
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

struct Edge {
    int s;
    int t;
    bool operator < (const Edge& tgt) const {
	if( s == tgt.s )
	    return t < tgt.t;
	return s < tgt.s;
    }
};



int Bridges_dfs(const vector<vector<Edge>>& g, vector<int>& order,
		vector<Edge>& ans, int u, int& cnt_v, Edge pre) {
    if( order[u] != -1 ) { //?¨????????????¨?????????
	return order[u];
    }
    // ??????????¨???????
    order[u] = cnt_v; cnt_v++;
    // ??£??\???????????????
    int min_v = order[u];
    for(auto e : g[u]) { // u ?????????????????????????????????????????????
	if( e.t == pre.s and e.s == pre.t ) continue; 
	min_v = min(Bridges_dfs(g, order, ans, e.t, cnt_v, e), min_v);
    }
    if( min_v == order[u] and pre.s != -1 ) {
	if(pre.s > pre.t) pre = Edge{pre.t,pre.s};
	ans.push_back(pre);
    }
    return min_v;
}

vector<Edge> Bridges(const vector<vector<Edge>>& g) {
    vector<Edge> ans;
    const int V = g.size();
    vector<int> order(V,-1);
    
    // 0??????DFS
    int cnt_v = 0;
    Edge tmp{-1,-1};
    Bridges_dfs(g, order, ans, 0, cnt_v, tmp);
    assert(cnt_v == V); // ??£???????????°?????????????????????
    return ans;
}

int main () {
    int V,E;
    cin >> V >> E;
    vector<vector<Edge>> g(V);
    for (int i=0; i<E; ++i) {
	int s,t;
	cin >> s >> t; 
	g[s].push_back(Edge{s,t});
	g[t].push_back(Edge{t,s});
    }
    vector<Edge> ans = Bridges(g);
    sort(ans.begin(), ans.end());
    for(auto& e : ans) {
	cout << e.s << " "<< e.t << endl;
    }

    return 0;
}
```