Problem:
Score : 300 points

Problem Statement

You are given an undirected connected graph with N vertices and M edges that does not contain self-loops and double edges.

The i-th edge (1 \leq i \leq M) connects Vertex a_i and Vertex b_i.

An edge whose removal disconnects the graph is called a bridge.

Find the number of the edges that are bridges among the M edges.

Notes

A self-loop is an edge i such that a_i=b_i (1 \leq i \leq M).

Double edges are a pair of edges i,j such that a_i=a_j and b_i=b_j (1 \leq i<j \leq M).

An undirected graph is said to be connected when there exists a path between every pair of vertices.

Constraints

2 \leq N \leq 50

N-1 \leq M \leq min(N(N−1)⁄2,50)

1 \leq a_i<b_i \leq N

The given graph does not contain self-loops and double edges.

The given graph is connected.

Input

Input is given from Standard Input in the following format:

N M
a_1 b_1
a_2 b_2
:
a_M b_M

Output

Print the number of the edges that are bridges among the M edges.

Sample Input 1

7 7
1 3
2 7
3 4
4 5
4 6
5 6
6 7

Sample Output 1

4

The figure below shows the given graph:

The edges shown in red are bridges. There are four of them.

Sample Input 2

3 3
1 2
1 3
2 3

Sample Output 2

0

It is possible that there is no bridge.

Sample Input 3

6 5
1 2
2 3
3 4
4 5
5 6

Sample Output 3

5

It is possible that every edge is a bridge.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<vector>
#include<algorithm>
#include<queue>
#include<cmath>
#include<iomanip>
#include<map>
#include<string>
using namespace std;

#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define vi vector<int>
#define vivi vector<vi>
const int INF = 1<<29;
typedef long long int ll;
int N, M;

int G[50][50];

int a[50], b[50];

void dfs(int i, bool visited[]){
    for(int j = 0;j < N;++j){
        if(!G[i][j])continue;
        if(visited[j])continue;
        visited[j] = true;
        dfs(j, visited);
    }
}

int main(){
    cin >> N >> M;
    rep(i, M){
        cin >> a[i] >> b[i];
        --a[i];--b[i];
        G[a[i]][b[i]] = 1;
        G[b[i]][a[i]] = 1;
    }
    int ans = 0;
    rep(i, M){
        bool visited[N];
        rep(i, N)visited[i] = false;
        visited[0] = true;
        G[a[i]][b[i]] = G[b[i]][a[i]] = 0;
        dfs(0, visited);
        bool flg = false;
        rep(j, N)if(!visited[j])flg = true;
        if(flg)ans += 1;
        G[a[i]][b[i]] = G[b[i]][a[i]] = 1;
    }
    cout << ans;
}