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;
vector<int> visited, active;
void dfs_visit(int root, const vector<vector<pair<int, int> > >& adj,
               vector<int>& colors, int parent) {
  visited[root] = 1;
  active[root] = 1;
  for (auto v : adj[root]) {
    colors[v.second] = active[v.first] ? 2 : 1;
    if (!visited[v.first]) dfs_visit(v.first, adj, colors, root);
  }
  active[root] = 0;
}
vector<int> dfs(int m, const vector<vector<pair<int, int> > >& adj) {
  vector<int> colors(m + 1);
  int n = adj.size() - 1;
  visited.resize(n + 1);
  active.resize(n + 1);
  for (int i = 1; i <= n; i++) {
    if (!visited[i]) dfs_visit(i, adj, colors, -1);
  }
  return colors;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  vector<vector<pair<int, int> > > adj(n + 1);
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back({v, i});
  }
  vector<int> colors = dfs(m, adj);
  int k = 1;
  for (auto& i : colors) k = max(k, i);
  cout << k << "\n";
  for (int i = 1; i <= m; i++) cout << colors[i] << " ";
  cout << "\n";
  return 0;
}
