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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <bits/stdc++.h>


#define f(i,n) for(int i=0;i<(n);i++)


#define inf (int)(3e18)
#define int long long
#define mod (int)(1000000007)

#define intt long long

using namespace std;

//Library
//モッドパウ
int modpow(int x, int y, int m = mod) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
            res %= m;
        }
        x = x * x % mod;
        y /= 2;
    }
    return res;
}

int mypow(int x, int y) {
    int res = 1;
    while (y) {
        if (y % 2) {
            res *= x;
        }
        x = x * x;
        y /= 2;
    }
    return res;
}
//is the number (x) a prime number?
bool prime(int x) {
    for (int i = 2; i * i <= x; i++) {
        if (!(x % i)) {
            return false;
        }
    }
    return true;
}

double kyori(pair<int, int> f, pair<int, int> s) {
    double ans = 0;
    double t = fabs(f.first - s.first);
    double y = fabs(f.second - s.second);
    ans = sqrt(t * t + y * y);
    return ans;
}

//saidai-kouyakusuu
int gcd(int x, int y) {
    if (!y) {
        return x;
    }
    return gcd(y, x % y);
}

//Union-Find Tree

class Union_Find {
    vector<int> par;
    vector<int> rankmy;
public:
    Union_Find(int size) {
        par = vector<int>(size);
        rankmy = vector<int>(size);
        for (int i = 0; i < size; i++) {
            par[i] = i;
        }
    }

    int find(int x) {
        if (par[x] == x) {
            return x;
        }
        return par[x] = find(par[x]);
    }

    void unite(int x, int y) {
        x = find(x);
        y = find(y);
        if (x == y) {
            return;
        }
        if (rankmy[x] < rankmy[y]) {
            par[x] = y;
        }
        else {
            par[y] = x;
            if (rankmy[x] == rankmy[y]) {
                rankmy[x]++;
            }
        }
    }

    bool same(int x, int y) {
        return find(x) == find(y);
    }

};

//Union-Find-End

//SegTree
template <class T>
class SegTree {
    int n;                       // 葉の数
    vector<T> data;              // データを格納するvector
    T def;                       // 初期値かつ単位元
    function<T(T, T)> operation; // 区間クエリで使う処理
    function<T(T, T)> update;    // 点更新で使う処理

    // 区間[a,b)の総和。ノードk=[l,r)に着目している。
    T _query(int a, int b, int k, int l, int r) {
        if (r <= a || b <= l) return def; // 交差しない
        if (a <= l && r <= b)
            return data[k]; // a,l,r,bの順で完全に含まれる
        else {
            T c1 = _query(a, b, 2 * k + 1, l, (l + r) / 2); // 左の子
            T c2 = _query(a, b, 2 * k + 2, (l + r) / 2, r); // 右の子
            return operation(c1, c2);
        }
    }

public:
    // _n:必要サイズ, _def:初期値かつ単位元, _operation:クエリ関数,
    // _update:更新関数
    SegTree(size_t _n, T _def, function<T(T, T)> _operation,
        function<T(T, T)> _update)
        : def(_def), operation(_operation), update(_update) {
        n = 1;
        while (n < _n) {
            n *= 2;
        }
        data = vector<T>(2 * n - 1, def);
    }

    // 場所i(0-indexed)の値をxで更新
    void change(int i, T x) {
        i += n - 1;
        data[i] = update(data[i], x);
        while (i > 0) {
            i = (i - 1) / 2;
            data[i] = operation(data[i * 2 + 1], data[i * 2 + 2]);
        }
    }

    // [a, b)の区間クエリを実行
    T query(int a, int b) {
        return _query(a, b, 0, 0, n);
    }

    // 添字でアクセス
    T operator[](int i) {
        return data[i + n - 1];
    }
};

#define R_MIN ([](long long a, long long b) { return min(a, b); })
#define R_MAX ([](long long a, long long b) { return max(a, b); })
#define R_SUM ([](long long a, long long b) { return a + b; })

#define NORMAL_UPDATE ([](long long a, long long b) { return b; })
#define ADD_UPDATE ([](long long a, long long b) { return a + b; })
#define MINUS_UPDATE ([](long long a, long long b) { return a - b; }
//Seg-Tree-End

//dfs
vector<int> v[100004];
bool went[100004];
void dfs(int x) {
    went[x] = true;
    for (int i = 0; i < (int)v[x].size(); i++) {
        if (!went[v[x][i]]) {
            dfs(v[x][i]);
        }
    }
}

//number of keta
int keta(int x) {
    int ans = 0;
    while (x) {
        x /= 10;
        ans++;
    }
    return ans;
}

//sum of keta
int ketasum(int x) {
    int ans = 0;
    while (x) {
        ans += x % 10;
        x /= 10;
    }
    return ans;
}

int lcm(int x, int y) {
    int ans = x / gcd(x, y) * y;
    return ans;
}
int twobeki(int x) {
    int ans = 0;
    while (1) {
        if (!(x & 1)) {
            ans++;
            x /= 2;
        }
        else {
            break;
        }
    }
    return ans;
}

int kai(int x, int y) {
    int res = 1;
    for (int i = x - y + 1; i <= x; i++) {
        res *= i; res %= mod;
    }
    return res;
}

int comb(int x, int y) {
    if (y > x)return 0;
    return kai(x, y) * modpow(kai(y, y), mod - 2) % mod;
}

int fukugen(vector<int> l) {
    int ans = 0;
    for (int i = 0; i < (int)l.size(); i++) {
        ans *= 10;
        ans += l[i];
    }
    return ans;
}
#define rep(i,n) for(int i=0;i<n;i++)

//Library-End
int n,m,a[54],b[54],ans;
signed main(){
    cin>>n>>m;
    rep(i, m){
        cin>>a[i]>>b[i];
        a[i]--;
        b[i]--;
    }
    rep(i, m){
        Union_Find uni(n);
        rep(j, m){
            if (i!=j) {
                uni.unite(a[j],b[j]);
            }
        }
        rep(j, n){
            if (!uni.same(0, j)) {
                ans++;
                break;
            }
        }
    }
    cout<<ans<<endl;
}