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 
n,m=[int(x) for x in input().split()]
dic={}
paint={}
cycle=False
edges=[]
for i in range(m):
    a,b=[int(x) for x in input().split()]
    edges.append((a,b))
    if a not in dic:
        dic[a]=set([b])
    else:
        dic[a].add(b)
inv=[False]*(n+1)
invite=[False]*(n+1)
def gcd(n):
    paint[n]=0
    invite[n]=True
    if n in dic:
        for item in dic[n]:
            if not invite[item]:
                if gcd(item):
                    return True
            else:
                if paint[item]==0:
                    return True
    paint[n]=1
    return False
for item in range(1,n+1):
    if not invite[item]:
        cycle=gcd(item) or cycle
arr=[]
if cycle:
    print(2)
    for item in edges:
        if item[0]<item[1]:
            arr.append(1)
        else:
            arr.append(2)
else:
    print(1)
    for item in edges:
        arr.append(1)
print(*arr)
    
