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;
int ans = 1;
int e[5005] = {0};
void dfs(int node, vector<vector<pair<int, int>>> &g, vector<int> &vis) {
  vis[node] = 1;
  for (auto v : g[node]) {
    if (vis[v.first] == 0) {
      dfs(v.first, g, vis);
    }
    if (vis[v.first] == 1) {
      ans = 2;
      e[v.second] = 2;
    }
  }
  vis[node] = 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  ans = 1;
  int u, v;
  vector<vector<pair<int, int>>> g(n);
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    g[u - 1].push_back({v - 1, i});
    e[i] = 1;
  }
  vector<int> vis(n);
  for (int i = 0; i < n; i++) vis[i] = 0;
  for (int i = 0; i < n; i++)
    if (vis[i] == 0) dfs(i, g, vis);
  cout << ans << endl;
  for (int i = 0; i < m; i++) {
    cout << e[i] << " ";
  }
  cout << endl;
  return 0;
}
