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;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
void tracer(istream_iterator<string> it) {}
template <typename T, typename... Args>
void tracer(istream_iterator<string> it, T a, Args... args) {
  cerr << *it << " = " << a << " | ";
  tracer(++it, args...);
}
const long long MOD = 1e9 + 7;
const double err = 1e-10;
const int N = 5e3 + 5;
vector<pair<int, int> > adj[N];
vector<char> color(N, 0);
vector<int> ord(N);
void dfs(int u) {
  color[u] = 1;
  for (auto prv : adj[u]) {
    int v = prv.first, idx = prv.second;
    if (color[v] == 0) {
      ord[idx] = 1;
      dfs(v);
    } else if (color[v] == 1) {
      ord[idx] = 2;
    } else {
      ord[idx] = 1;
    }
  }
  color[u] = 2;
}
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(make_pair(v, i));
  }
  for (int i = 1; i < n + 1; i++) {
    if (color[i] == 0) {
      dfs(i);
    }
  }
  int mx = *max_element(ord.begin(), ord.begin() + m);
  cout << mx << "\n";
  for (int i = 0; i < m; i++) {
    cout << ord[i] << " ";
  }
  cout << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T = 1;
  for (int i = 1; i <= T; i++) {
    solve();
  }
  return 0;
}