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 = input().split()
n = int(n)
m = int(m)
ady = [[] for _ in range(n + 1)]
color = [0] * (n + 1) #every vertice is WHITE to the begin
oneColor = True
sol = [0] * m

def dfs_visit(v): # 0 is WHITE, 1 is GRAY, 2 is BLACK
    global sol, ady, oneColor, color
    
    color[v] = 1 #v become GRAY
    
    for par in ady[v]:
        u, index = par[0], par[1]
        if(color[u] == 0): #tree edge
            dfs_visit(u)
            sol[index] = 1 #(v, u) will be blue
        elif(color[u] == 2): #cross edge or forward edge
            sol[index] = 1 #(v, u) will be blue
        else: #back edge, we have a cycle!!!!
            sol[index] = 2 #(v, u) will be red
            oneColor = False

    color[v] = 2 #v become in BLACK    

for i in range(m):
    n1, n2 = input().split()
    n1 = int(n1)
    n2 = int(n2)
    ady[n1].append((n2, i))

for v in range(1, n + 1):
    if(color[v] == 0):
        dfs_visit(v)

if(oneColor):
    print(1)
else:
    print(2)

res = str("")
for i in range(0, m):
    res += str(sol[i]) + str(" ")

print(res)
    