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 
def DFS(graph, status, v):
    status[v] = 1
    for u in graph[v]:
        if status[u] == 1:
            return True
        if status[u] == 0:
            if DFS(graph, status, u):
                return True
    status[v] = 2


def containsCycle(graph):
    n = len(graph)
    status = [0] * n
    for v in range(n):
        if status[v] != 0:
            continue
        if DFS(graph, status, v):
            return True
    return False


nm = input().split()
n = int(nm[0])
m = int(nm[1])

graph = [set() for _ in range(n)]
edges = []
for _ in range(m):
    edge = input().split()
    u = int(edge[0]) - 1
    v = int(edge[1]) - 1
    edges.append((u, v))

    graph[u].add(v)

if (not containsCycle(graph)):
    print(1)
    answer = [1]*m
    print(*answer)
else:
    print(2)
    answer = []
    for e in edges:
        if e[0] > e[1]:
            answer.append(1)
        else:
            answer.append(2)
    print(*answer)
