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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val (n, m) = readLine()!!.split(' ').map { it.toInt() }
    val edges = mutableListOf<Pair<Int, Int>>()
    val g = Array(n + 1) { mutableListOf<Int>() }
    repeat(m) {
        val (a, b) = readLine()!!.split(' ').map { it.toInt() }
        edges.add(a to b)
        g[a].add(b)
        g[b].add(a)
    }

    fun dfs(i: Int): Boolean {
        val ignore = edges[i].let { setOf(it.first, it.second) }
        val visited = BooleanArray(n + 1) { false }

        fun rec(u: Int): Int {
            visited[u] = true
            var cnt = 1
            for (v in g[u]) {
                if (visited[v]) continue
                if (ignore.contains(u) && ignore.contains(v)) continue
                cnt += rec(v)
            }
            return cnt
        }

        return rec(1) < n
    }

    println((0 until m).count { dfs(it) })
}