Problem: 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 

Solution (in PYTHON3):

Solution:
import sys
import collections
import functools

n,m  = map(int, input().strip().split())
graph = collections.defaultdict(list)
edges = []
for i in range(m):
    u,v = map(int, input().strip().split())
    graph[u].append([v,i])
    edges.append((u,v))

ans = [1]*m
visited = set()

def dfs(u, seen):
    seen.add(u)
    visited.add(u)
    for v, i in graph[u]:
        if v in seen:
            ans[i] = 2
        elif v not in visited:
            ans[i] = 1
            dfs(v, seen)
    seen.remove(u)

seen = set()

for u, v in edges:
    if u in visited: continue
    dfs(u, seen)

print(max(ans))
print(' '.join(str(i) for i in ans))