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;
map<pair<long long, long long>, long long> order;
void dfs(vector<vector<long long>> &edges, long long s,
         vector<long long> &visited, vector<long long> &color) {
  visited[s] = 1;
  for (long long i = 0; i < edges[s].size(); i++) {
    if (visited[edges[s][i]] == 0) {
      color[order[{s, edges[s][i]}]] = 1;
      dfs(edges, edges[s][i], visited, color);
    } else if (visited[edges[s][i]] == 1) {
      color[order[{s, edges[s][i]}]] = 2;
    } else {
      color[order[{s, edges[s][i]}]] = 1;
    }
  }
  visited[s] = 2;
}
void mymain() {
  long long n, m;
  cin >> n >> m;
  vector<vector<long long>> edges(n + 1, vector<long long>());
  for (long long i = 0; i < m; i++) {
    long long u, v;
    cin >> u >> v;
    order[{u, v}] = i;
    edges[u].push_back(v);
  }
  vector<long long> visited(n + 1, 0);
  vector<long long> color(m, 0);
  for (long long i = 1; i <= n; i++) {
    if (visited[i] == 0) {
      dfs(edges, i, visited, color);
    }
  }
  set<long long> s;
  for (long long i : color) s.insert(i);
  cout << s.size() << '\n';
  for (long long i : color) cout << i << " ";
  cout << '\n';
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t = 1;
  for (long long tt = 0; tt < t; tt++) {
    mymain();
  }
  return 0;
}
