// 边双连通分量

// https://www.luogu.com.cn/problem/P8436

#include <bits/stdc++.h>

using namespace std;

vector<vector<int>> g, edccs;

vector<int> dfn, low;

vector<bool> instk;

stack<int> stk;

int tot;

// tarjan算法求边双连通分量(eDCCs)

void tarjan(int u, int fa) {

dfn[u] = low[u] = tot++;

stk.push(u);

instk[u] = true;

// 是否访问到反边

bool fa_edge = false;

for (auto v : g[u]) {

if (dfn[v] == -1) {

tarjan(v, u);

low[u] = min(low[u], low[v]);

// 找到割边

// if (low[v] > dfn[v]);

} else if (instk[v] && (v != fa || fa_edge))

low[u] = min(low[u], dfn[v]);

else

fa_edge = true;

}

if (dfn[u] != low[u]) return;

edccs.push_back({});

int v;

do {

v = stk.top();

stk.pop();

edccs.back().push_back(v);

instk[v] = false;

} while (v != u);

}

int main() {

int n, m;

cin >> n >> m;

g.resize(n);

dfn.resize(n, -1);

low.resize(n, -1);

instk.resize(n);

for (int i = 0; i < m; ++i) {

int u, v;

cin >> u >> v;

u--;

v--;

g[u].push_back(v);

g[v].push_back(u);

}

for (int i = 0; i < n; ++i)

if (dfn[i] == -1) tarjan(i, -1);

cout << edccs.size() << "\n";

for (auto& edcc : edccs) {

cout << edcc.size() << " ";

for (auto u : edcc) cout << u + 1 << " ";

cout << "\n";

}

return 0;

}