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 <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

typedef pair<int, int> Pair;

int n, m;
int low[100000];
int ord[100000];
bool used[100000];
vector<Pair> ans;
vector<int> es[100000];

void dfs(int v, int p, int &idx) {
    used[v] = true;
    low[v] = ord[v] = idx++;

    for (int i=0; i<es[v].size(); i++) {
        int u = es[v][i];
        if (!used[u]) {
            dfs(u, v, idx);
            low[v] = min(low[v], low[u]);
            if (ord[v] < low[u]) {
                int s = min(u, v);
                int t = max(u, v);
                ans.push_back(Pair(s, t));
            }
        } else if (u != p) {
            low[v] = min(low[v], ord[u]);
        }
    }
}
    

int main() {
    scanf("%d%d", &n, &m);
    for (int i=0; i<m; i++) {
        int a, b;
        scanf("%d%d", &a, &b);
        es[a].push_back(b);
        es[b].push_back(a);
    }

    int _ = 0;
    dfs(0, 0, _);

    sort(ans.begin(), ans.end());
    for (int i=0; i<ans.size(); i++) {
        printf("%d %d\n", ans[i].first, ans[i].second);
    }
}