Problem: 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 

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;

public class CF1217D extends PrintWriter {
	CF1217D() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1217D o = new CF1217D(); o.main(); o.flush();
	}

	int[] next, jj; int l_ = 1;
	int link(int l, int j) { next[l_] = l; jj[l_] = j; return l_++; }
	int[] ao, cc;
	boolean[] instack, visited;
	void init(int n, int m) {
		next = new int[1 + m];
		jj = new int[1 + m];
		cc = new int[m];
		ao = new int[n];
		instack = new boolean[n];
		visited = new boolean[n];
	}
	boolean dfs(int i) {
		if (visited[i])
			return instack[i];
		visited[i] = instack[i] = true;
		for (int l = ao[i]; l != 0; l = next[l]) {
			int j = jj[l];
			if (dfs(j))
				return true;
		}
		instack[i] = false;
		return false;
	}
	void main() {
		int n = sc.nextInt();
		int m = sc.nextInt();
		init(n, m);
		for (int h = 0; h < m; h++) {
			int i = sc.nextInt() - 1;
			int j = sc.nextInt() - 1;
			ao[i] = link(ao[i], j);
			cc[h] = i < j ? 1 : 2;
		}
		int k = 1;
		for (int i = 0; i < n; i++)
			if (!visited[i] && dfs(i)) {
				k = 2;
				break;
			}
		println(k);
		if (k == 1)
			Arrays.fill(cc, 1);
		for (int h = 0; h < m; h++)
			print(cc[h] + " ");
		println();
	}
}