Problem:
Score : 100 points

Problem Statement

You are given a directed graph with N vertices and M edges, not necessarily simple. The i-th edge is oriented from the vertex a_i to the vertex b_i.
Divide this graph into strongly connected components and print them in their topological order.

Constraints

1 \leq N \leq 500,000

1 \leq M \leq 500,000

0 \leq a_i, b_i < N

Input

Input is given from Standard Input in the following format:

N M
a_0 b_0
a_1 b_1
:
a_{M - 1} b_{M - 1}

Output

Print 1+K lines, where K is the number of strongly connected components.
Print K on the first line.
Print the information of each strongly connected component in next K lines in the following format, where l is the number of vertices in the strongly connected component and v_i is the index of the vertex in it.

l v_0 v_1 ... v_{l-1}

Here, for each edge (a_i, b_i), b_i should not appear in earlier line than a_i.

If there are multiple correct output, print any of them.

Sample Input 1

6 7
1 4
5 2
3 0
5 5
4 1
0 3
4 2

Sample Output 1

4
1 5
2 4 1
1 2
2 3 0

Solution in Python (3.8.2) (judge verdict: Accepted)

# verify-helper: PROBLEM https://judge.yosupo.jp/problem/scc
import sys

input = sys.stdin.buffer.readline
sys.setrecursionlimit(6 * 10 ** 5)


class StronglyConnectedComponets:
    def __init__(self, n: int) -> None:
        self.n = n
        self.edges = [[] for _ in range(n)]
        self.rev_edeges = [[] for _ in range(n)]
        self.vs = []
        self.order = [0] * n
        self.used = [False] * n

    def add_edge(self, from_v: int, to_v: int) -> None:
        self.edges[from_v].append(to_v)
        self.rev_edeges[to_v].append(from_v)

    def dfs(self, v: int) -> None:
        self.used[v] = True
        for child in self.edges[v]:
            if not self.used[child]:
                self.dfs(child)
        self.vs.append(v)

    def rdfs(self, v: int, k: int) -> None:
        self.used[v] = True
        self.order[v] = k
        for child in self.rev_edeges[v]:
            if not self.used[child]:
                self.rdfs(child, k)

    def run(self) -> int:
        self.used = [False] * self.n
        self.vs.clear()
        for v in range(self.n):
            if not self.used[v]:
                self.dfs(v)
        self.used = [False] * self.n
        k = 0
        for v in reversed(self.vs):
            if not self.used[v]:
                self.rdfs(v, k)
                k += 1
        return k


def main() -> None:
    N, M = map(int, input().split())
    scc = StronglyConnectedComponets(N)
    for _ in range(M):
        u, v = map(int, input().split())
        scc.add_edge(u, v)
    scc.run()
    ans_size = max(scc.order) + 1
    ans = [[] for _ in range(ans_size)]
    for i, c in enumerate(scc.order):
        ans[c].append(i)
    print(ans_size)
    for row in ans:
        print(len(row), *row)


if __name__ == "__main__":
    main()