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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

import qualified Data.IntMap as IM
import qualified Data.Vector as V
main=print.solve.map(map read.words).lines=<<getContents
solve::[[Int]]->Int
solve([n,m]:abs)=fst(f(-1)0 IM.empty)-1 where
  f s t v=(c+if p==w IM.! t then 1 else 0,w)where
    p=IM.size v
    (c,w)=foldr h(0,IM.insert t p v)$g V.! t
    h i a@(c,v)
      |i==s=a
      |IM.member i v=(c,IM.adjust(min(v IM.! i))t v)
      |otherwise=let(d,w)=f t i v in(c+d,IM.adjust(min(w IM.! i))t w)
  g=V.accum(flip(:))(V.replicate n[])$concatMap(\[a,b]->[(a,b),(b,a)])$map(map pred)abs