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 
from collections import defaultdict,deque
import sys
import bisect
input=sys.stdin.readline

n,m=map(int,input().split())
graph=[tuple(map(int,input().split())) for i in range(m)]
e=[[] for i in range(n+1)]

e_in=[0]*(n+1)
e_out=[[] for i in range(n+1)]
for x,y in graph:
    e_in[y]+=1
    e_out[x].append(y)

    
queue=deque()
#print(e_in)
for i in range(1,n+1):
    if e_in[i]==0:
        queue.append(i)

top=[]
while queue:
    x=queue.popleft()
    top.append(x)
    for child in e_out[x]:
        e_in[child]-=1
        if e_in[child]==0:
            queue.append(child)
#print(top,e_in)
if len(top)==n:
    print(1)
    print(*[1]*(m))
else:
    print(2)
    for x,y in graph:
        if x>y:
            print(1,end=' ')
        else:
            print(2,end=' ')
            
        
        
    
        
    
    
