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] = list(map(int,input().split()))
edges = [[] for _ in range(N+1)]
edge_in = []

for _ in range(M):
    [u,v] = list(map(int,input().split()))
    edge_in.append([u,v])
    edges[u].append(v)

seen = [False for _ in range(N+1)]
visited = [False for _ in range(N+1)]

def bfs(node):
    visited[node] = True
    seen[node] = True
    for v in edges[node]:
        if not seen[v] and visited[v]:
            continue
        if seen[v] or bfs(v):
            return True
    seen[node] = False
    return False

hasCycle = False
for i in range(1,N+1):
    if visited[i]:
        continue
    if bfs(i):
        hasCycle = True
        break

if not hasCycle:
    print(1)
    print(" ".join(["1" for _ in range(M)]))
else:
    print(2)
    print(" ".join(["1" if u < v else "2" for (u,v) in edge_in]))
