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;
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
const int N = 5005;
vector<pair<long long, long long> > v[N];
long long vis[N];
long long color[N];
long long k;
void dfs(long long node) {
  vis[node] = 1;
  for (auto i : v[node]) {
    if (vis[i.first] == 1) {
      color[i.second] = 2;
      k = 2;
      continue;
    } else {
      color[i.second] = 1;
      if (!vis[i.first]) dfs(i.first);
    }
  }
  vis[node] = 2;
}
void solve() {
  long long n, m, x, y;
  cin >> n >> m;
  for (long long i = 1; i <= m; i++) {
    cin >> x >> y;
    v[x].push_back({y, i});
  }
  k = 1;
  for (long long i = 1; i <= n; i++)
    if (!vis[i]) dfs(i);
  cout << k << '\n';
  for (long long i = 1; i <= m; i++) cout << color[i] << " ";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int TESTS = 1;
  while (TESTS--) {
    solve();
  }
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}