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;
struct hash_pair {
  template <class T1, class T2>
  size_t operator()(const pair<T1, T2>& p) const {
    auto hash1 = hash<T1>{}(p.first);
    auto hash2 = hash<T2>{}(p.second);
    return hash1 ^ hash2;
  }
};
long long t;
long long m = 1e9 + 7;
long long arr[5005] = {1}, chk[5005];
vector<long long> v[5005];
map<pair<long long, long long>, long long> mp;
long long res = 1;
void dfs(long long a) {
  chk[a] = 1;
  for (long long i = 0; i < v[a].size(); i++) {
    long long b = v[a][i];
    if (chk[b] == 1) {
      long long d = mp[{a, b}];
      res = 2;
      arr[d] = 2;
    } else if (chk[b] == 2)
      continue;
    else
      dfs(b);
  }
  chk[a] = 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long a, b, c, d, e, f, g, i, j;
  cin >> a >> b;
  for (i = 1; i <= b; i++) arr[i] = 1;
  f = 0;
  for (i = 1; i <= b; i++) {
    cin >> c >> d;
    v[c].push_back(d);
    mp[{c, d}] = i;
  }
  for (i = 1; i <= a; i++) {
    if (chk[i] == 0) {
      dfs(i);
    }
  }
  cout << res << endl;
  for (i = 1; i <= b; i++) {
    if (i != 1) cout << ' ';
    cout << arr[i];
  }
  return 0;
}
