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 Rust (1.15.1) (judge verdict: Accepted)

macro_rules! input {
    ($t : ty) => {{
        use std::io::BufRead;
        let tmp_handle = std::io::stdin();
        let mut handle = tmp_handle.lock();
        let mut buf = String::new();
        handle.read_line(&mut buf).unwrap();
        let tmp_var: $t = buf.trim().parse().unwrap(); 
        buf.clear();
        tmp_var
    }};
    ($t : ty, $size : expr) => {{
        use std::io::BufRead;
        let tmp_handle = std::io::stdin();
        let mut handle = tmp_handle.lock();
        let mut buf = String::new();
        handle.read_line(&mut buf).unwrap();
        let tmp_var: $t = buf.trim().split_whitespace()
            .map(|a| a.parse().unwrap()).collect();
        buf.clear();
        tmp_var
    }};
}
#[derive(Debug)]
pub struct UnionFind {
    parents: Vec<usize>,
    rank: Vec<usize>,
}
impl UnionFind {
    pub fn new(n: usize) -> Self {
        let mut parents: Vec<usize> = Vec::with_capacity(n);
        for i in 0..n {
            parents.push(i);
        }
        let rank = vec![1; n];
        UnionFind {
            parents: parents,
            rank: rank,
        }
    }
    pub fn find(&mut self, x: usize) -> usize {
        if self.parents[x] == x {
            x
        } else {
            let par = self.parents[x];
            let root = self.find(par);
            self.parents[x] = root;
            root
        }
    }
    // union x and y
    // ルートを付け替えるだけで，新たに付け加えられた木の子供の親は変わらないことに注意
    // unionしたあとも find を呼ぶ必要がある．
    pub fn union(&mut self, x: usize, y: usize) {
        let x_root = self.find(x);
        let y_root = self.find(y);
        // if x_root == y_root, do nothing
        if x_root == y_root { return (); }
        if self.rank[x_root] > self.rank[y_root] {
            // x <--- y
            self.parents[y_root] = x_root;
        } else {
            // y <--- x
            self.parents[x_root] = y_root;
            if self.rank[x_root] == self.rank[y_root] {
                self.rank[y_root] += 1;
            }
        }
    }
}
fn main() {
    let config = input!(Vec<usize>, 2);
    use std::collections::VecDeque;
    let mut edges = VecDeque::with_capacity(config[1]);
    for _ in 0..config[1] {
        let edge = input!(Vec<usize>, 2);
        edges.push_back(edge);
    }
    let mut count = 0;
    for _ in 0..config[1] {
        let one_edge = edges.pop_front().unwrap();
        let mut uf = UnionFind::new(config[0]);
        for edge in &edges {
            uf.union(edge[0]-1, edge[1]-1);
        }
        use std::collections::HashSet;
        let mut vs = HashSet::with_capacity(config[0]);
        for v in 0..config[0] {
            vs.insert(uf.find(v));
        }
        if vs.len() == 1 {
            // one_edge is not a bridge
        } else if vs.len() == 2 {
            // one_edge is a bridge
            count += 1;
        } else {
            panic!("my misunderstanding...");
        }
        edges.push_back(one_edge);
    }
    println!("{}", count);
}