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;
vector<long long> g[5011], w[5011];
long long ans[5011];
long long vis[5011];
long long cyc;
long long dfs(long long u, long long po) {
  vis[u] = 1;
  for (long long i = 0; i < g[u].size(); ++i) {
    long long v = g[u][i], wt = w[u][i];
    if (ans[wt] == 0) {
      ans[wt] = 3 - po;
      dfs(v, 3 - po);
    }
  }
  return 0;
}
long long in[5011], out[5011];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m, i;
  cin >> n >> m;
  vector<pair<long long, long long> > vp;
  for (i = 1; i <= m; ++i) {
    long long x, y;
    cin >> x >> y;
    g[x].push_back(y);
    in[y]++;
    vp.push_back({x, y});
  }
  queue<long long> q;
  for (i = 1; i <= n; ++i)
    if (in[i] == 0) q.push(i);
  cyc = 0;
  while (!q.empty()) {
    long long c = q.front();
    q.pop();
    ++cyc;
    for (auto v : g[c]) {
      in[v]--;
      if (in[v] == 0) q.push(v);
    }
  }
  if (cyc == n) {
    cout << "1\n";
    for (i = 1; i <= m; ++i) cout << 1 << " ";
    cout << "\n";
    return 0;
  }
  cout << "2\n";
  for (i = 1; i <= m; ++i)
    cout << 1 + (vp[i - 1].first < vp[i - 1].second) << " ";
  cout << "\n";
  return 0;
}