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 int N = 5000;
bool viz[1 + N];
bool used[1 + N];
vector<pair<int, int>> gr[1 + N];
int ans[1 + N];
void dfs(int node) {
  used[node] = true;
  viz[node] = true;
  for (auto son : gr[node]) {
    if (used[son.first]) {
      ans[son.second] = 2;
      continue;
    }
    if (viz[son.first]) continue;
    dfs(son.first);
  }
  used[node] = false;
}
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    gr[x].push_back({y, i});
  }
  for (int i = 1; i <= m; i++) ans[i] = 1;
  for (int i = 1; i <= n; i++)
    if (!viz[i]) dfs(i);
  int k = 0;
  for (int i = 1; i <= m; i++) k = max(k, ans[i]);
  cout << k << "\n";
  for (int i = 1; i <= m; i++) cout << ans[i] << " ";
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
  return 0;
}