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 <string.h>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <vector>

using namespace std;

const int kMaxV = 10010;

struct Edge {
  int u;
  int v;

  Edge(int u, int v): u(u), v(v) {}

  bool operator < (const Edge& rhs) const {
    if (u != rhs.u) {
      return u < rhs.u;
    } else {
      return v < rhs.v;
    }
  }
};

int v, e, source = 0, dfs_clock = 0;
vector<int> g[kMaxV];
int pre[kMaxV] = {0};
int low[kMaxV] = {0};
int parent[kMaxV];
vector<Edge> bridges;

void DFS(int u, int father) {
  int child_num = 0;
  ++dfs_clock;
  pre[u] = dfs_clock;
  low[u] = dfs_clock;
  for (int v: g[u]) {
    if (!pre[v]) {
      ++child_num;
      parent[v] = u;
      DFS(v, u);
      low[u] = min(low[u], low[v]);
      if (low[v] > pre[u]) {
        bridges.emplace_back(min(u, v), max(u, v));
      }
    } else if (v != father) {
      low[u] = min(low[u], pre[v]);
    }
  }
}

void Bridge() {
  memset(parent, source, sizeof(parent));
  DFS(source, -1);
  sort(bridges.begin(), bridges.end());
}

int main() {
  scanf("%d %d", &v, &e);
  for (int i = 0; i < e; ++i) {
    int s, t;
    scanf("%d %d", &s, &t);
    g[s].push_back(t);
    g[t].push_back(s);
  }

  Bridge();
  for (auto b: bridges) {
    printf("%d %d\n", min(b.u, b.v), max(b.u, b.v));
  }

  return 0;
}