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>
#pragma GCC optimize("Ofast")
using namespace std;
template <typename A1>
void prn(A1&& arg) {
  cout << arg << '\n';
}
template <typename A1, typename... A>
void prn(A1&& arg, A&&... args) {
  cout << arg << ' ';
  prn(args...);
}
vector<pair<long long, long long> > v[5005];
bool b[5005], c[5005];
long long e[5005];
void dfs(long long x) {
  b[x] = 1, c[x] = 1;
  for (pair<long long, long long> p : v[x]) {
    if (!b[p.first])
      dfs(p.first);
    else if (c[p.first])
      e[p.second] = 1;
  }
  c[x] = 0;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long x, y, n, m;
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    cin >> x >> y;
    v[x].emplace_back(y, i);
  }
  for (long long i = 1; i <= n; i++)
    if (!b[i]) dfs(i);
  long long ans = 0;
  for (long long i = 0; i < m; i++) e[i]++, ans = max(ans, e[i]);
  prn(ans);
  for (long long i = 0; i < m; i++) cout << e[i] << ' ';
  cout << '\n';
}
