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 <bits/stdc++.h>
#define rep(i,n) for(int i=0; i<(n); i++)
#define FOR(i,a,n) for(int i=(a); i<(n); i++)
#define all(v) v.begin(), v.end()
#define fi first
#define se second
using namespace std;
using ll = long long;
using P = pair<int ,int>;
const int INF = 1e9;
const int MOD = 1e9+7;

int n, m;
vector<bool> seen;
vector<vector<P>> G;

void dfs(int i, int v) {
    seen[v]=true;
    for (const auto& nv: G[v]) {
        if (i==nv.se) continue;
        if (seen[nv.fi]) continue;
        dfs(i, nv.fi);
    }
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> n >> m;
    G.resize(n);
    rep(i, m) {
        int x, y; cin >> x >> y;
        x--; y--;
        G[x].emplace_back(y, i);
        G[y].emplace_back(x, i);
    }
    int ans = 0;
    rep(i, m) {
        seen.assign(n, false);
        dfs(i, 0);
        rep(i, n) {
            if(!seen[i]) {ans++; break;}
        }
    }

    cout << ans << endl;
}