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;
const long long md = 2e7 + 3;
const long long MN = 2e7 + 3;
long long n;
vector<vector<int> > g;
vector<char> cl;
vector<long long> p;
int cycle_st, cycle_end;
bool dfs(int v) {
  cl[v] = 1;
  for (size_t i = 0; i < g[v].size(); ++i) {
    int to = g[v][i];
    if (cl[to] == 0) {
      p[to] = v;
      if (dfs(to)) return true;
    } else if (cl[to] == 1) {
      cycle_end = v;
      cycle_st = to;
      return true;
    }
  }
  cl[v] = 2;
  return false;
}
int main() {
  long long j, m, n, k, q, r;
  cin >> n >> m;
  vector<pair<long, long> > c;
  for (int i = 0; i < int(n); i++) g.push_back({});
  for (int i = 0; i < int(m); i++) {
    cin >> k >> q;
    c.push_back({k, q});
    g[k - 1].push_back(q - 1);
  }
  p.assign(n, -1);
  cl.assign(n, 0);
  cycle_st = -1;
  for (int i = 0; i < n; ++i)
    if (dfs(i)) break;
  if (cycle_st == -1) {
    cout << 1 << endl;
    for (int i = 0; i < int(m); i++) cout << 1 << " ";
  } else {
    cout << 2 << endl;
    for (int i = 0; i < int(m); i++) {
      if (c[i].first > c[i].second)
        cout << 1 << " ";
      else
        cout << 2 << " ";
    }
  }
}
namespace {
auto fast_io = [] {
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  return 0;
}();
}