You are given a directed graph with n vertices and m directed edges without self-loops or multiple edges.

Let's denote the k-coloring of a digraph as following: you color each edge in one of k colors. The k-coloring is good if and only if there no cycle formed by edges of same color.

Find a good k-coloring of given digraph with minimum possible k.

Input

The first line contains two integers n and m (2 ≤ n ≤ 5000, 1 ≤ m ≤ 5000) — the number of vertices and edges in the digraph, respectively.

Next m lines contain description of edges — one per line. Each edge is a pair of integers u and v (1 ≤ u, v ≤ n, u ≠ v) — there is directed edge from u to v in the graph.

It is guaranteed that each ordered pair (u, v) appears in the list of edges at most once.

Output

In the first line print single integer k — the number of used colors in a good k-coloring of given graph.

In the second line print m integers c_1, c_2, ..., c_m (1 ≤ c_i ≤ k), where c_i is a color of the i-th edge (in order as they are given in the input).

If there are multiple answers print any of them (you still have to minimize k).

Examples

Input


4 5
1 2
1 3
3 4
2 4
1 4


Output


1
1 1 1 1 1 


Input


3 3
1 2
2 3
3 1


Output


2
1 1 2 
#!python3

from collections import deque, Counter
import array
from itertools import combinations, permutations
from math import sqrt
import unittest


def read_int():
    return int(input().strip())


def read_int_array():
    return [int(i) for i in input().strip().split(' ')]

######################################################

vn, en = read_int_array()
al = [[] for _ in range(vn)]  # adjacency list
def adj(v):
    return al[v]
itoe = [None for _ in range(en)]  # index to edge
for eid in range(en):  # eid - edge id
    v, w = read_int_array()
    v -= 1
    w -= 1
    al[v] += [w]
    itoe[eid] = (v, w)

marked = set()
stack = set()
etoc = {}  # edge to color
def dfs(v):  # vertex
    if v in marked:
        return
    marked.add(v)
    stack.add(v)
    hasbackedge = False
    for w in adj(v):
        if w in stack:  # back edge
            hasbackedge = True
            etoc[(v, w)] = 2
            continue
        if w in marked:
            continue
        if dfs(w):
            hasbackedge = True
    stack.remove(v)
    return hasbackedge

hasbackedge = False
for v in range(vn):
    if v not in marked:
        if dfs(v):
            hasbackedge = True

print(2 if hasbackedge else 1)
for ei in range(en):
    v, w = itoe[ei]
    c = etoc.get((v,w), 1)
    print(c, end=' ')





