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)

// inputに入力データ全体が入る
function Main(input) {
    var N = input[0].split(" ")[0]-0;
    var M = input[0].split(" ")[1]-0;
    var list = {};
    for(var i=1;i<=M;i++){
        var a = input[i].split(" ")[0]-0;
        var b = input[i].split(" ")[1]-0;
        list[a] = list[a]||{};
        list[b] = list[b]||{};
        list[a][b] = i;
        list[b][a] = i;
    }
    
    var ans = 0;
    for(var i=1;i<=M;i++){
        if(check(list, i)!=N)ans++;
    }
    
    console.log(ans);
}


function check(list, i){
    var stack = [1];
    var checked = {};
    while(stack.length>0){
        var now = stack.splice(0,1)[0];
        if(checked[now]) continue;
        checked[now] = true;
        var next = Object.keys(list[now]);
        next.forEach(nxt=>{
            if(list[now][nxt]!=i){
                stack.push(nxt);
            }
        });
    }
    return Object.keys(checked).length;
}


Main(require("fs").readFileSync("/dev/stdin", "utf8").trim().split("\n"));