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;
void solve() {
  int n, m;
  cin >> n >> m;
  vector<vector<int>> g(n);
  vector<int> co(m);
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    x--;
    y--;
    g[x].emplace_back(y);
    co[i] = x > y ? 1 : 2;
  }
  int sta = 1, fin = 2;
  vector<int> trk(n);
  function<bool(int)> dfs = [&](int u) {
    trk[u] = sta;
    for (int v : g[u]) {
      if (trk[v] < sta) {
        if (dfs(v)) return true;
      } else if (trk[v] < fin) {
        return true;
      }
    }
    trk[u] = fin;
    return false;
  };
  bool has_cycle = false;
  for (int i = 0; i < n; i++) {
    if (!trk[i]) has_cycle |= dfs(i);
  }
  int res = 2;
  if (!has_cycle) res = 1, fill(co.begin(), co.end(), 1);
  cout << res << "\n";
  for (auto& x : co) {
    cout << x << ' ';
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  solve();
  cout << endl;
}
