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 = map(int,raw_input().split(" "))

edges = [[] for i in range(5005)]

for i in range(m):
    a,b = map(int,raw_input().split(" "))
    edges[a].append((b,i))

cnt     = 0
visited = [0]*5005
flag    = True
colors = [1]*m

cn = 0
k = 1
while True:
  while visited[k] != 0:
        k += 1
  if k >n:
    break
  nodes   = [k]
  visited[k] = 1

  while  nodes:
    node    = nodes.pop()
    visited[node] = 1
    cnt     = 0
    first   = True
    for nd,idx in edges[node]:
        if not visited[nd]:
            if first:
                nodes.append(node)
                first = False
            nodes.append(nd)
        elif visited[nd] == 2:
            continue
        elif visited[nd] == 1:
            colors[idx] = 2
    if first :
        visited[node] = 2

print len(set(colors))
print " ".join(map(str,colors))
