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;
long long n, m;
vector<vector<pair<long long, long long>>> g;
vector<long long> col, ans;
bool ok = true;
void dfs(long long v) {
  col[v] = 1;
  for (auto it : g[v]) {
    long long to = it.first, ind = it.second;
    if (!col[to]) {
      dfs(to);
      ans[ind] = 1;
    } else if (col[to] == 2)
      ans[ind] = 1;
    else {
      ok = false;
      ans[ind] = 2;
    }
  }
  col[v] = 2;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  g.assign(n, vector<pair<long long, long long>>());
  for (long long i = 0; i < m; i++) {
    long long u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back({v, i});
  }
  col.resize(n);
  ans.resize(m);
  for (long long v = 0; v < n; v++)
    if (!col[v]) dfs(v);
  if (ok)
    cout << 1 << '\n';
  else
    cout << 2 << '\n';
  for (long long i = 0; i < m; i++) cout << ans[i] << ' ';
  cout << '\n';
  return 0;
}
