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 "fmt"

func main() {
	var N, M, cnt int
	fmt.Scan(&N, &M)
	a := make([]int, M)
	b := make([]int, M)
	for i := 0; i < M; i++ {
		fmt.Scan(&a[i], &b[i])
	}
	for i := 0; i < M; i++ {
		un := UnionFind{}
		un.init(N)
		for j := 0; j < M; j++ {
			if i != j {
				un.unite(a[j], b[j])
			}
		}
		var c int
		for j := 0; j < N; j++ {
			if j == un.par[j] {
				c++
			}
		}
		if 1 < c {
			cnt++
		}
	}
	fmt.Println(cnt)
}

// UnionFind
type UnionFind struct {
	par []int
	cnt int
}

func (u *UnionFind) init(n int) {
	for i := 0; i <= n; i++ {
		u.par = append(u.par, i)
	}
}
func (u *UnionFind) root(x int) int {
	if u.par[x] == x {
		return x
	} else {
		u.par[x] = u.root(u.par[x])
		return u.par[x]
	}
}
func (u *UnionFind) same(x, y int) bool {
	return u.root(x) == u.root(y)
}
func (u *UnionFind) unite(x, y int) {
	x = u.root(x)
	y = u.root(y)
	if x == y {
		return
	}
	u.par[x] = y
}