#include <bits/stdc++.h>
using namespace std;
/* typedef */
typedef long long ll;
typedef pair<int, int> pii;
/* constant */
const int INF = 1 << 30;
const ll LINF = 1LL << 50;
const int NIL = -1;
const int MAX = 100005;
const int mod = 1000000007;
const double pi = 3.141592653589;
/* global variables */
int N, M;
int cnt = 0;
vector<int> ord(MAX, NIL), lowLink(MAX);
vector<vector<int> > G(MAX, vector<int>());
/* function */
void dfs(int from, int parent) {
    ord[from] = lowLink[from] = cnt++;
    for (int to : G[from]) {
        // DFS木の親を訪れようとしたら何もしない
        if (to == parent) continue;
        if (ord[to] == NIL) {
            // nodeが使われていないならdfs
            dfs(to, from);
            // dfsが戻ってきたらlowlinkの更新
            lowLink[from] = min(lowLink[from], lowLink[to]);
        }
        // 使われている頂点まだ使ってない辺でを訪れようとした場合
        else {
            // これが後退辺となるのでusedEは0のままにしておく
            // lowlinkの更新
            lowLink[from] = min(lowLink[from], ord[to]);
        }
    }
    return ;
}
void bridge(vector<pii> edges) {
    int from, to;
    for (pii edge : edges) {
        from = edge.first, to = edge.second;
        if (ord[from] < lowLink[to] || ord[to] < lowLink[from])
            cout << from << ' ' << to << '\n';
    }
}
/* main */
int main(){
    cin >> N >> M;
    int u, v;
    vector<pii> edges;
    for (int i = 0; i < M; i++) {
        cin >> u >> v;
        if (u > v) swap(u, v);
        G[u].push_back(v);
        G[v].push_back(u);
        edges.push_back(pii(u, v));
    }
    dfs(0, -1);
    sort(edges.begin(), edges.end());
    bridge(edges);
}


### Prompt

Create a solution in Cpp for 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 */
typedef long long ll;
typedef pair<int, int> pii;
/* constant */
const int INF = 1 << 30;
const ll LINF = 1LL << 50;
const int NIL = -1;
const int MAX = 100005;
const int mod = 1000000007;
const double pi = 3.141592653589;
/* global variables */
int N, M;
int cnt = 0;
vector<int> ord(MAX, NIL), lowLink(MAX);
vector<vector<int> > G(MAX, vector<int>());
/* function */
void dfs(int from, int parent) {
    ord[from] = lowLink[from] = cnt++;
    for (int to : G[from]) {
        // DFS木の親を訪れようとしたら何もしない
        if (to == parent) continue;
        if (ord[to] == NIL) {
            // nodeが使われていないならdfs
            dfs(to, from);
            // dfsが戻ってきたらlowlinkの更新
            lowLink[from] = min(lowLink[from], lowLink[to]);
        }
        // 使われている頂点まだ使ってない辺でを訪れようとした場合
        else {
            // これが後退辺となるのでusedEは0のままにしておく
            // lowlinkの更新
            lowLink[from] = min(lowLink[from], ord[to]);
        }
    }
    return ;
}
void bridge(vector<pii> edges) {
    int from, to;
    for (pii edge : edges) {
        from = edge.first, to = edge.second;
        if (ord[from] < lowLink[to] || ord[to] < lowLink[from])
            cout << from << ' ' << to << '\n';
    }
}
/* main */
int main(){
    cin >> N >> M;
    int u, v;
    vector<pii> edges;
    for (int i = 0; i < M; i++) {
        cin >> u >> v;
        if (u > v) swap(u, v);
        G[u].push_back(v);
        G[v].push_back(u);
        edges.push_back(pii(u, v));
    }
    dfs(0, -1);
    sort(edges.begin(), edges.end());
    bridge(edges);
}


```