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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
string to_string(string s) { return '"' + s + '"'; }
string to_string(const char* s) { return to_string((string)s); }
string to_string(bool b) { return (b ? "true" : "false"); }
template <typename A, typename B>
string to_string(pair<A, B> p) {
  return "(" + to_string(p.first) + ", " + to_string(p.second) + ")";
}
template <typename A>
string to_string(A v) {
  bool first = true;
  string res = "{";
  for (const auto& x : v) {
    if (!first) {
      res += ", ";
    }
    first = false;
    res += to_string(x);
  }
  res += "}";
  return res;
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << " " << to_string(H);
  debug_out(T...);
}
vector<long long int> g[6000];
map<pair<long long int, long long int>, long long int> col;
vector<long long int> vis(6000, 0);
vector<long long int> curr(6000, 0);
void dfs(long long int root) {
  vis[root] = 1;
  curr[root] = 1;
  for (auto i : g[root]) {
    if (curr[i] == 1)
      col[{root, i}] = 2;
    else
      col[{root, i}] = 1;
    if (vis[i] == 0) dfs(i);
  }
  curr[root] = 0;
}
signed main() {
  std::ios::sync_with_stdio(false);
  long long int n, m;
  cin >> n >> m;
  vector<pair<long long int, long long int> > edges;
  for (long long int i = 0; i < m; i++) {
    long long int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    edges.push_back({x, y});
  }
  for (long long int i = 1; i <= n; i++)
    if (vis[i] == 0) dfs(i);
  long long int k = 1;
  for (auto i : col)
    if (i.second == 2) k = 2;
  cout << k << "\n";
  for (auto i : edges) cout << col[i] << " ";
}