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>) = yorukatsu18d()

fun yorukatsu18d() {
    val (n, m) = readLine()!!.split(' ').map { it.toInt() }
    val edges = (1..m).map {
        readLine()!!.split(' ').map { it.toInt() - 1 }.let { Edge(it[0], it[1]) }
    }.toSet()

    val answer = edges.count { edge ->
        val other = edges - edge
        val uf = UnionFind(n)
        other.forEach { uf.unite(it.from, it.to) }
        (0 until n).map { uf.root(it) }.distinct().size != 1
    }

    println(answer)
}

private data class Edge(val from: Int, val to: Int)

private class UnionFind(initSize: Int) {
    private val parent = IntArray(initSize) { it }
    private val size = LongArray(initSize) { 1L }

    fun root(x: Int): Int {
        if (x == parent[x]) return x

        val xr = root(parent[x])
        parent[x] = xr
        return xr
    }

    fun unite(x: Int, y: Int) {
        val xr = root(x)
        val yr = root(y)
        if (xr == yr) return

        if (size[xr] >= size[yr]) {
            size[xr] += size[yr]
            parent[yr] = xr
        } else {
            size[yr] += size[xr]
            parent[xr] = yr
        }
    }

    fun isSame(x: Int, y: Int): Boolean = root(x) == root(y)

    fun size(x: Int): Long = size[root(x)]
}