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, input().split())
numedge = {}
graph = {}
for v in range(1, n + 1):
    graph[v] = set()
for i in range(m):
    a, b = map(int, input().split())
    numedge[(a, b)] = i
    graph[a].add(b)
visit = {}
par = {}
s, f = -1, -1


def dfs(v):
    global s, f
    visit[v] = 1
    for u in graph[v]:
        if u not in visit:
            par[u] = v
            dfs(u)
        elif visit[u] == 1:
            s = u
            f = v
            break
    visit[v] = -1


for v in graph:
    dfs(v)
if s == f == -1:
    print(1)
    print('1 ' * m)
else:
    print(2)
    ans = [0] * m
    for x in graph:
        for y in graph[x]:
            ans[numedge[(x, y)]] = (x > y) + 1
    print(*ans)
