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
sys.setrecursionlimit(10**7)
def MI(): return map(int,sys.stdin.readline().rstrip().split())


class SCC:
    def __init__(self,N):
        self.N = N
        self.graph = [[] for _ in range(N)]
        self.graph_rev = [[] for _ in range(N)]
        self.flag = [False]*N

    def add_edge(self,start,end):
        if start == end:
            return
        self.graph[start].append(end)
        self.graph_rev[end].append(start)

    def dfs(self,node,graph):
        self.flag[node] = True
        for n in graph[node]:
            if self.flag[n]:
                continue
            self.dfs(n,graph)
        self.order.append(node)

    def first_dfs(self):
        self.flag = [False]*self.N
        self.order = []
        for i in range(self.N):
            if self.flag[i] == False:
                self.dfs(i,self.graph)

    def second_dfs(self):
        self.flag = [False]*self.N
        self.ans = []
        for n in reversed(self.order):
            if self.flag[n]:
                continue
            self.flag[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


N,M = MI()
SCC = SCC(N)

for _ in range(M):
    a,b = MI()
    SCC.add_edge(a,b)

ANS = SCC.scc()

print(len(ANS))
for s in ANS:
    print(*([len(s)]+s))