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 Lua (5.3.2) (judge verdict: Accepted)

n, m = io.read('n'), io.read('n')
e = {}
for i = 1, n do e[i] = {} end
for i = 1, m do
    local u, v = io.read('n'), io.read('n')
    e[u][#e[u] + 1] = v
    e[v][#e[v] + 1] = u
end

dfn, low = {}, {}
tot = 0
bridge_count = 0

dfs = function (u, p)
    tot = tot + 1
    dfn[u] = tot
    low[u] = tot
    for _, v in ipairs(e[u]) do if v ~= p then
        if dfn[v] == nil then
            dfs(v, u)
            low[u] = math.min(low[u], low[v])
        else
            low[u] = math.min(low[u], dfn[v])
        end
    end end
    if p ~= -1 and low[u] > dfn[p] then
        bridge_count = bridge_count + 1
    end
end

for i = 1, n do if dfn[i] == nil then dfs(i, -1) end end

print(bridge_count)