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:

from collections import defaultdict
import sys

input = sys.stdin.buffer.readline

def scc(V, E):
    # g = defaultdict(list)
    g = [[] for _ in range(len(V))]
    # rg = defaultdict(list)
    rg = [[] for _ in range(len(V))]
    for u, v in E:
        g[u].append(v)
        rg[v].append(u)


    o = []
    done = set()
    for u in V:
        if u in done:
            continue

        s = [(u, True)]
        while s:
            u, f = s.pop()
            if f:
                if u in done:
                    continue
                done.add(u)
                s.append((u, False))
                for v in g[u]:
                    if v in done:
                        continue
                    s.append((v, True))
            else:
                o.append(u)

    done = set()
    ans = []
    while o:
        u = o.pop()
        if u in done:
            continue

        s = [u]
        vv = [u]
        done.add(u)
        while s:
            u = s.pop()
            for v in rg[u]:
                if v in done:
                    continue
                vv.append(v)
                s.append(v)
                done.add(v)
        ans.append(vv)


    return ans


def atcoder_practice2_g():
    N, M = map(int, input().split())

    E = []
    for _ in range(M):
        a, b = map(int, input().split())
        E.append((a, b))
    V = list(range(N))
    vv = scc(V, E)
    print(len(vv))
    for r in vv:
        print(len(r), *r)


def test():
    g = {
        1: [2],
        5: [2],
        2: [4, 3],
        3: [4, 7, 6],
        4: [5, 8],
        6: [9],
        7: [8],
        8: [9],
        9: [7, 10],
        10: []
    }

    V = list(range(1, 11))
    E = []
    for u, e in g.items():
        for v in e:
            E.append((u, v))

    vv = scc(V, E)
    for r in vv:
        print(r)


if __name__ == "__main__":
    # test()
    atcoder_practice2_g()