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 
vis = []
ans = []
nodes = []


def dfs(A):
	global vis, ans, nodes
	vis[A] = 1
	for (B,index) in nodes[A]:
		if ans[index] != 0:
			continue
		if vis[B] == 1:
			ans[index] = 2
		else:
			ans[index] = 1
			dfs(B)
			
	vis[A] = 2

def main():
	global vis, ans, nodes
	n, m = map(int, input().strip().split())
	vis = [-1] * (n + 10)
	ans = [0] * (m + 10)
	nodes = [[] for i in range(n + 10)]
	for i in range(m):
		a, b = map(int, input().strip().split())
		nodes[a].append((b,i))
	for i in range(1, n + 1):
		if vis[i] == -1:
			dfs(i)
	if 2 in ans:
		print(2)
	else:
		print(1)
	for i in range(m):
		print(ans[i])

if __name__ == '__main__':
	main()