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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
        "bufio"
        "fmt"
        "os"
        "strconv"
)

var si = bufio.NewScanner(os.Stdin)

func readInt() int {
        si.Scan()
        r, _ := strconv.Atoi(si.Text())
        return r
}

type UnionTree struct {
        id     int
        parent *UnionTree
}

func (p *UnionTree) Root() *UnionTree {
        if p.parent == nil {
                return p
        }
        root := p.parent.Root()
        if p.parent != root {
                p.parent = root
        }
        return p.parent
}
func (p *UnionTree) unite(q *UnionTree) {
        pRoot := p.Root()
        qRoot := q.Root()
        if pRoot == qRoot{
                return
        }
        if pRoot.id > qRoot.id {
                p.parent = qRoot
        } else {
                q.parent = pRoot
        }
}
func (p *UnionTree) same(q *UnionTree) bool {
        return p.Root() == q.Root()
}
func main() {
        si.Split(bufio.ScanWords)
        si.Buffer(make([]byte, 0), 10000000)
        N, M := readInt(), readInt()
        e := make([][2]int, M)
        for i := 0; i < M; i++ {
                a, b := readInt()-1, readInt()-1
                e[i] = [2]int{a, b}
        }
        ret := 0
        for i := 0; i < M; i++ {
                t := make([]*UnionTree, N)
                for i := 0; i < N; i++ {
                        t[i] = &UnionTree{i, nil}
                }
                for j := 0; j < M; j++ {
                        if j != i {
                                t[e[j][0]].unite(t[e[j][1]])
                        }
                }
                if !t[e[i][0]].same(t[e[i][1]]) {
                        ret += 1
                }
        }
        fmt.Println(ret)
}