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 v5.12) (judge verdict: Accepted)

log = console.log;

var stdin = require("fs").readFileSync("/dev/stdin", "UTF-8");
var lines = stdin.split("\n");
var N = lines[0].split(" ").map(Number)[0];
var M = lines[0].split(" ").map(Number)[1];
var A = lines
  .slice(1)
  .filter(v => v !== "")
  .map(v => v.split(" ").map(Number));

var nLists = [];

for (var i = 0; i < M; i++) {
  nLists[i] = [];
  A.forEach((nm, A_i) => {
    if (A_i === i) return;
    var n = nm[0];
    var m = nm[1];
    if (nLists[i][m] === undefined) {
      nLists[i][m] = [n];
    } else {
      nLists[i][m].push(n);
    }
    if (nLists[i][n] === undefined) {
      nLists[i][n] = [m];
    } else {
      nLists[i][n].push(m);
    }
  });
}

function dfs(list, start, visited) {
  visited[start] = 1;
  list[start].forEach(point => {
    if (!visited[point]) dfs(list, point, visited);
  });
}

count = 0;
nLists.forEach(list => {
  var visited = [0];
  var firstPoint = list.findIndex(v=> v !== undefined)
  dfs(list, firstPoint, visited);
  if (visited.reduce((a,b)=>a+b) !== N) count += 1;
});

log(count);