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 
import sys
input = sys.stdin.readline
def dfs(cur_node, childs, vis, cur_dfs):
    if cur_node in cur_dfs:
        return True

    if vis[cur_node]:
        return False

    vis[cur_node] = True
    cur_dfs.add(cur_node)
    for ele in childs[cur_node]:
        if dfs(ele, childs, vis, cur_dfs):
            return True

    cur_dfs.remove(cur_node)
    return False

n, m = map(int, input().split())
childs = [[] for i in range(n+1)]
has_dad = [False] * (n+1)
vis = [False] * (n+1)
ans2 = []
for i in range(m):
    x1, x2 = map(int, input().split())
    ans2.append(str((x1 < x2) + 1))
    childs[x1].append(x2)
    has_dad[x2] = True

has_cycle = False
for i in range(1, n+1):
    if not has_dad[i] and dfs(i, childs, vis, set()):
        has_cycle = True
        break

for i in range(1, n+1):
    if has_dad[i] and not vis[i]:
        has_cycle = True
        break

if has_cycle:
    print(2)
    print(' '.join(ans2))
else:
    print(1)
    print(' '.join(['1']*m))
