Problem: 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.

Example

Input

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


Output

4
1 5
2 4 1
1 2
2 3 0

Solution (in PYTHON3):

Solution:
import sys
input = sys.stdin.readline
sys.setrecursionlimit(1000000)

class SCC:
    def __init__(self, n):
        self.n = n
        self.graph = [[] for _ in range(n)]
        self.graph_rev = [[] for _ in range(n)]
        self.already = [False]*n
    
    def add_edge(self, fr, to):
        if fr == to:
            return
        self.graph[fr].append(to)
        self.graph_rev[to].append(fr)
    
    def dfs(self, node, graph):
        self.already[node] = True
        for n in graph[node]:
            if self.already[n]:
                continue
            self.dfs(n, graph)
        self.order.append(node)
    
    def first_dfs(self):
        self.already = [False]*self.n
        self.order = []
        for i in range(self.n):
            if self.already[i] == False:
                self.dfs(i, self.graph)
    
    def second_dfs(self):
        self.already = [False]*self.n
        self.ans = []
        for n in reversed(self.order):
            if self.already[n]:
                continue
            self.already[n] = True
            self.order = []
            self.dfs(n, self.graph_rev)
            self.order.reverse()
            self.ans.append(self.order)

    def scc(self):
        self.first_dfs()
        self.second_dfs()
        return self.ans


def main():
    n, m = map(int, input().split())
    scc = SCC(n)

    for i in range(m):
        a, b = map(int, input().split())
        scc.add_edge(a, b)
    
    scc_list = scc.scc()

    print(len(scc_list))
    for s in scc_list:
        print(len(s), end=" ")
        print(*s)

    
if __name__ == "__main__":
    main()