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>
using namespace std;
typedef long long int ll;
#define rep(i, n) for (int i = 0; i < (int)n; i++)
int N, M;
int Ai[50];
int Bi[50];
int tree[51];

int parent(int n) {
    if (tree[n] == n) {
        return n;
    }
    int p = parent(tree[n]);
    return tree[n] = p;
}
//i番目が橋かどうかチェック
bool isBridge(int idx) {
    //idx番目を除去したときに島が独立になるか調べる
    rep(i, N) tree[i] = i;

    rep(i, M) {
        if (i == idx) {
            continue;
        }
        int a = parent(Ai[i]);
        int b = parent(Bi[i]);
        if (a == b) {
            continue;
        }
        tree[b] = a;
    }
    //全部が同一グループにある＝橋でない
    int p0 = parent(0);
    rep (i, N) {
        if (parent(i) != p0) {
            return true;
        }
    }
    return false;
}


int main()
{
    cin >> N >> M;
    rep(i, M) {
        int a, b;
        cin >> a >> b;
        Ai[i] = a - 1;
        Bi[i] = b - 1;
    }
    int count = 0;
    rep (i, M) {
        if (isBridge(i)) {
            count++;
        }
    }
    cout << count << endl;
    return 0;
}