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 <bits/stdc++.h>
using namespace std;

const int maxn = 1e5 + 5;
int mark[maxn];
int par[maxn];
int hig[maxn];
int zir[maxn];
int dp[maxn];
int minDpKids[maxn];
int minHig[maxn];

vector<int> adj[maxn];
vector< pair<int,int> > output;

void dfs (int v) {
    mark[v] = 1;
    zir[v] = 1;

    for (int u : adj[v] ) {
        if ( !mark[u] ) {
            par[u] = v;
            hig[u] = hig[v] + 1;
            dfs(u);
            zir[v] += zir[u];
            dp[v] = min (dp[v] , dp[u]);
        }
        else if ( u != par[v] && hig[u] < hig[v]) {
            dp[v] = min(hig[u] , dp[v]);
        }
    }

    if (dp[v] >= hig[v] && dp[v] + hig[v] != 0){

        pair<int,int> outy (min (par[v] , v) , max (par[v] , v)) ;
        output.push_back(outy);
    }
}

int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    int n , m;
    cin >> n >> m;

    for(int i=1; i<n; i++)
        dp[i] = 100005;

    for (int i = 0; i<m; i++) {
        int u, v;
        cin >> u >> v;

        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    dfs(0);

    sort (output.begin() , output.end());

    for(int i = 0; i < output.size(); i++)
        cout << output[i].first << ' ' << output[i].second << endl;

    return 0;
}

/*

4 4
0 1
0 2
1 2
2 3
----------
5 4
0 1
1 2
2 3
3 4
----------
6 5
0 1
0 2
0 3
1 4
4 5

*/