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("O3")
using namespace std;
long long mod = 1e9 + 7;
vector<long long> g[5003];
bool vis[5003], f;
vector<pair<long long, long long> > ed;
void dfs(long long i, long long p) {
  vis[i] = 1;
  for (auto j : g[i]) {
    if (j == p) f = 1;
    if (!vis[j]) dfs(j, p);
  }
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  long long i, j, x, y;
  f = 0;
  for (i = 1; i <= m; ++i) {
    cin >> x >> y;
    ed.push_back({x, y});
    g[x].push_back(y);
  }
  for (i = 1; i <= n and f == 0; ++i) {
    fill(vis, vis + 5003, 0);
    dfs(i, i);
  }
  if (f) {
    cout << "2\n";
    for (auto it : ed)
      if (it.first < it.second)
        cout << "1 ";
      else
        cout << "2 ";
  } else {
    cout << "1\n";
    for (auto it : ed) cout << "1 ";
  }
  return 0;
}
