#include <bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
template<typename A, typename B> inline bool chmax(A &a, B b) { if (a<b) { a=b; return 1; } return 0; }
template<typename A, typename B> inline bool chmin(A &a, B b) { if (a>b) { a=b; return 1; } return 0; }
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const ll INF = 1ll<<29;
const ll MOD = 1000000007;
const double EPS  = 1e-9;
const bool debug = 0;
//---------------------------------//

int V, E;
vector<int> g[112345];
vector<int> depth, low;
set<pii> bridge;

int dfs(int u, int p, int d) {
	depth[u] = d;
	low[u] = d;
	
	for (int v : g[u]) {
		if (v == p) continue;
		
		if (depth[v] == -1) {
			chmin(low[u], dfs(v, u, d + 1));
			if (depth[v] == low[v]) bridge.insert(pii(min(u, v), max(u, v)));
		}
		else chmin(low[u], low[v]);
	}
	
	return low[u];
}


int main() {
	cin >> V >> E;
	REP(i, E) {
		int s, t;
		scanf("%d %d", &s, &t);
		g[s].push_back(t);
		g[t].push_back(s);
	}
	
	depth.resize(V, -1);
	low.resize(V, -1);
	
	dfs(0, -1, 0);
	
	for (pii now : bridge) printf("%d %d\n", now.first, now.second);
	
	return 0;
}
### Prompt

In CPP, your task is to solve 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;
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
template<typename A, typename B> inline bool chmax(A &a, B b) { if (a<b) { a=b; return 1; } return 0; }
template<typename A, typename B> inline bool chmin(A &a, B b) { if (a>b) { a=b; return 1; } return 0; }
typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
const ll INF = 1ll<<29;
const ll MOD = 1000000007;
const double EPS  = 1e-9;
const bool debug = 0;
//---------------------------------//

int V, E;
vector<int> g[112345];
vector<int> depth, low;
set<pii> bridge;

int dfs(int u, int p, int d) {
	depth[u] = d;
	low[u] = d;
	
	for (int v : g[u]) {
		if (v == p) continue;
		
		if (depth[v] == -1) {
			chmin(low[u], dfs(v, u, d + 1));
			if (depth[v] == low[v]) bridge.insert(pii(min(u, v), max(u, v)));
		}
		else chmin(low[u], low[v]);
	}
	
	return low[u];
}


int main() {
	cin >> V >> E;
	REP(i, E) {
		int s, t;
		scanf("%d %d", &s, &t);
		g[s].push_back(t);
		g[t].push_back(s);
	}
	
	depth.resize(V, -1);
	low.resize(V, -1);
	
	dfs(0, -1, 0);
	
	for (pii now : bridge) printf("%d %d\n", now.first, now.second);
	
	return 0;
}
```