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 <algorithm>
#include <iostream>
#include <limits>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <vector>
using namespace std;
class edge {
   public:
    int t;
    int id;
    edge(int to, int i) : t(to), id(i) {}
    edge() {}
};
int n = 0, m = 0;
vector<vector<edge>> g;
vector<int> pre;
vector<int> ll;
vector<pair<int, int>> res;
int tt = 0;
void input() {
    cin >> n >> m;
    int id = 1, M = m;
    g = vector<vector<edge>>(n, vector<edge>());
    pre = vector<int>(n, -1);
    ll = vector<int>(n, -1);
    while (M--) {
        int a = 0, b = 0;
        cin >> a >> b;
        g[a].push_back({b, ++id});
        g[b].push_back({a, id});
    }
}
int dfs(int i, int p, int id) {
    if (pre[i] != -1) {
        ll[p] = min(ll[p], pre[i]);
        return ll[p];
    }
    pre[i] = ++tt;
    ll[i] = pre[i];
    for (auto v : g[i]) {
        if (v.id == id) {
            continue;
        }
        if (dfs(v.t, i, v.id) < 0) {
            ll[i] = min(ll[i], ll[v.t]);
            if (ll[v.t] == pre[v.t]) res.push_back({i, v.t});
        }
    }
    return -1;
}
void compute() {
    for (int i = 0; i < n; i++) {
        if (pre[i] == -1) {
            dfs(i, i, m + 2);
        }
    }
    for (auto &r : res) {
        if (r.first > r.second) {
            swap(r.first, r.second);
        }
    }
    sort(res.begin(), res.end());

    for (auto b : res) {
        cout << b.first << " " << b.second << endl;
    }
}

int main() {
    input();
    compute();
    return 0;
}