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;
template <class T>
istream &operator>>(istream &in, vector<T> &arr) {
  for (auto &x : arr) in >> x;
  return in;
}
long long n, m;
const long long N = 1e6 + 100;
vector<pair<long long, long long>> g[N];
long long col[N];
bool cyc;
long long res[N];
void dfs(long long r) {
  col[r] = 1;
  for (auto x : g[r]) {
    long long to = x.first, ind = x.second;
    if (col[to] == 0) {
      res[ind] = 1;
      dfs(to);
    } else if (col[to] == 1) {
      res[ind] = 2;
      cyc = true;
    } else {
      res[ind] = 1;
    }
  }
  col[r] = 2;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  for (long long i = 0; i < m; ++i) {
    long long u, v;
    cin >> u >> v;
    --u, --v;
    g[u].push_back({v, i});
  }
  for (long long i = 0; i < n; ++i) {
    if (col[i] == 0) {
      dfs(i);
    }
  }
  cout << (cyc ? 2 : 1) << '\n';
  for (long long i = 0; i < m; ++i) {
    cout << res[i] << ' ';
  }
}