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 
import sys
sys.setrecursionlimit(10**9)
def dfs(a):
    global adj,vis,st
    j=0
    vis[a]=1
    for i in adj[a][:]:
        if vis[i[0]]==0:
            dfs(i[0])
            adj[a][j][1]=1
        elif vis[i[0]]==2:
            adj[a][j][1]=1
        else:
            st=2
            adj[a][j][1]=2
        j+=1
    vis[a]=2
 
n,m=map(int,input().split())
it=[]
adj=[[] for i in range(n)]
#print(adj)
for _ in range(m):
    a,b=map(int,input().split())
    adj[a-1].append([b-1,1,_])
    #print(adj)
vis=[0]*n
st=1
for ii in range(n):
    if vis[ii]==0:
        dfs(ii)
print(st)
ans=[0]*m
for i in range(n):
    for j in adj[i]:
      #  print(j[2],j[1])
        ans[j[2]]=j[1]
print(*ans)
 
