Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
class DColoringEdges {
 public:
  void solve(std::istream& cin, std::ostream& cout) {
    int n, m;
    cin >> n >> m;
    vector<int> g[n + 1];
    int indg[n + 1];
    vector<pair<int, int>> edg;
    for (int i = 0; i <= n; i++) indg[i] = 0;
    for (int i = 0; i < m; i++) {
      int u, v;
      cin >> u >> v;
      edg.push_back(make_pair(u, v));
      g[u].push_back(v);
      indg[v]++;
    }
    queue<int> q;
    vector<int> topo;
    for (int i = 1; i <= n; i++) {
      if (indg[i] == 0) {
        q.push(i);
      }
    }
    while (!q.empty()) {
      int nd = q.front();
      q.pop();
      topo.push_back(nd);
      for (int x : g[nd]) {
        indg[x]--;
        if (indg[x] == 0) q.push(x);
      }
    }
    if (topo.size() == n) {
      cout << 1 << endl;
      for (int i = 0; i < m; i++) cout << 1 << " ";
      cout << endl;
    } else {
      cout << 2 << endl;
      for (pair<int, int> e : edg) {
        if (e.first < e.second)
          cout << 1 << " ";
        else
          cout << 2 << " ";
      }
      cout << endl;
    }
  }
};
int main() {
  DColoringEdges solver;
  std::ios::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}