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 JavaScript (Node.js 12.16.1) (judge verdict: Accepted)

function Main(input) {
	input = input.trim().split("\n").map(function(x) { return x.split(" ")});    
	let N = parseInt(input[0][0], 10);
	let M = parseInt(input[0][1], 10);
	let line;
	let ans = 0;
	for (skip = 0; skip < M; skip++){
		let uf = new UnionFind(N);
		for (let i = 0; i < M; i++){
			if (i === skip) continue;
			line = input[i + 1].map(e => parseInt(e, 10));
			uf.union(line[0] - 1, line[1] - 1);
		}
		if (uf.size(0) !== N) ans ++;
	}	
	console.log(ans);
}

function UnionFind (n) {
	this.parents = [];
	for (let i = 0; i < n; i++) {
		this.parents.push(-1);
	}
}

UnionFind.prototype.find = function (x) {
	if (this.parents[x] < 0) return x;
	return this.parents[x] = this.find(this.parents[x]);	
}

UnionFind.prototype.union = function (x, y) {
	x = this.find(x);
	y = this.find(y);
	if (x === y) return;
	if (this.parents[x] > this.parents[y]) [x, y] = [y, x];
	
	this.parents[x] += this.parents[y];
	this.parents[y] = x;
}

UnionFind.prototype.size = function (x) {
	x = this.find(x);
	while (x >= 0) {
		x = this.parents[x];
	}
	return x * -1;
}

Main(require("fs").readFileSync("/dev/stdin", "utf8"));