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 
#include <bits/stdc++.h>
using namespace std;
bool dfs(int v, vector<int> &flag, vector<vector<int> > &g) {
  if (flag[v] == 2) return false;
  if (flag[v] == 1) return true;
  flag[v] = 1;
  for (int z = 0; z < int(g[v].size()); z++) {
    int j = g[v][z];
    if (dfs(j, flag, g)) return true;
  }
  flag[v] = 2;
  return false;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  vector<pair<int, int> > e(m);
  vector<vector<int> > g(n + 1);
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &e[i].first, &e[i].second);
    g[e[i].first].push_back(e[i].second);
  }
  vector<int> flag(n + 1, 0);
  bool has_cycle = false;
  for (int i = 1; i <= n; i++) has_cycle |= dfs(i, flag, g);
  vector<int> ans(m);
  for (int i = 0; i < m; i++) ans[i] = (e[i].first < e[i].second) ? 1 : 2;
  int k = 2;
  if (!has_cycle) fill(ans.begin(), ans.end(), k = 1);
  printf("%d\n", k);
  for (int i = 0; i < m; i++) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}
