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 collections import defaultdict
ans = defaultdict(lambda : 1)
flag = 0
def dfs(i):

    global flag
    vis[i] = 1

    for j in hash[i]:
        if not vis[j]:
            dfs(j)
        else:
            if vis[j] == 1:
                flag = 1

                ans[(i,j)] = 2


    vis[i] = 2



n,m = map(int,input().split())

hash = defaultdict(list)

par = [0]+[i+1 for i in range(n)]



for i in range(m):
    a,b = map(int,input().split())

    hash[a].append(b)

    ans[(a,b)] = 1

vis = [0]*(n+1)

for i in range(n):
    if vis[i] == 0:
        dfs(i)
if flag:
    print(2)
else:
    print(1)
for i in ans:
    print(ans[i],end = ' ')





