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())
 
AE = []
E = [[] for _ in range(n)]
di = [0]*n
do = [0]*n
ok = True
 
def DFS(u):
    global ok
    
    di[u] = True
    for v in E[u]:
        if di[v]:
            if not do[v]:
                ok = False
            continue
        DFS(v)
    do[u] = True
 
for _ in range(m):
    u, v = map(int, input().split())
    u -= 1
    v -= 1
    
    AE.append((u,v) )
    E[u].append(v)
 
for i in range(n):
    if not di[i]:
        DFS(i)

if ok:
    print(1)
    for _ in range(m):
        print(1, end=' ')
else:
    print(2)
    for u,v in AE:
        print(int(u < v) + 1, end=' ')