#include <bits/stdc++.h>

using namespace std;
//此题是P8436的弱化版本，弱化了记录每个边双中点有哪些:vector<int> belong[N];
const int N = 500010, M = 2000010 * 2;

int n, m;

int dfn[N], low[N], ts, stk[N], top;
vector<int> belong[N]; //边双中有哪些原始点

int id[N], dcc_cnt; //原始点x属于哪个边双连通分量，dcc_cnt指边双连通分量个数
int is_bridge[M];   //记录哪些边是割边

//链式前向星
int e[M], h[N], idx, w[M], ne[M];
void add(int a, int b, int c = 0) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}

//边双连通分量
void tarjan(int u, int fa) {
    dfn[u] = low[u] = ++ts;
    stk[++top] = u;
    for (int i = h[u]; ~i; i = ne[i]) {
        int j = e[i];
        if (!dfn[j]) {
            tarjan(j, i);
            low[u] = min(low[u], low[j]);
            if (dfn[u] < low[j]) is_bridge[i] = is_bridge[i ^ 1] = true; //记录割边
        } else if (i != (fa ^ 1))
            low[u] = min(low[u], dfn[j]);
    }

    if (dfn[u] == low[u]) {
        ++dcc_cnt; //边双数量+1
        int x;
        do {
            x = stk[top--];
            id[x] = dcc_cnt;              // 记录点与边双关系
            belong[dcc_cnt].push_back(x); // 记录边双中有哪些点
        } while (x != u);
    }
}

int main() {
    scanf("%d %d", &n, &m);
    memset(h, -1, sizeof h);

    for (int i = 0; i < m; i++) {
        int a, b;
        scanf("%d %d", &a, &b);
        add(a, b), add(b, a);
    }

    for (int i = 1; i <= n; i++)
        if (!dfn[i]) tarjan(i, -1);

    //输出边双个数
    printf("%d\n", dcc_cnt);

    // for (int i = 1; i <= dcc_cnt; i++) {
    //     //此边双中点的数量
    //     printf("%d ", belong[i].size());
    //     //此边双中都有哪些点
    //     for (int j = 0; j < belong[i].size(); j++)
    //         printf("%d ", belong[i][j]);
    //     puts("");
    // }

    return 0;
}