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 sys import stdin, stdout

time=0
k=1
def dfs(node: int, graph):
    
    stack = []
    global time
    global k
    stack.append(node)
    while stack:
        s = stack.pop()
        if d[s] == -1:
            d[s] = time
            time += 1
            stack.append(s)

            for v, i in graph[s]:
                if d[v] == -1:
                    colour[i] = "1"
                    stack.append(v)
                else:
                    if f[v] == -1:
                        k=2
                        colour[i] = "2"
                    else:
                        colour[i] = "1"
        elif f[s] == -1:
            f[s] = time
            time += 1


n, m = [int(x) for x in stdin.readline().split()]
graph = [[] for _ in range(n)]
for i in range(m):
    u, v = stdin.readline().split()
    u, v = (int(u)-1, int(v)-1)
    graph[u].append((v, i))

d = [-1]*n
f = [-1]*n
pi = [-1]*n
colour = ["1"]*m

for i in range(n):
    if d[i]==-1:
        dfs(i, graph)
stdout.write(f'{k}\n')
col = ' '.join(colour)
col +='\n'
stdout.write(col)