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;
int k = 1;
int colour[5005];
int from[5005];
int seen[5005];
pair<int, int> edges[5005];
vector<pair<int, int> > graph[5005];
bool isAncestor(int curr, int prev) {
  if (curr == prev) return true;
  if (from[curr] == curr) return false;
  return isAncestor(from[curr], prev);
}
void dfs(int root, int prev) {
  if (seen[root]) return;
  seen[root] = 1;
  from[root] = prev;
  for (pair<int, int> i : graph[root]) {
    if (seen[i.first]) {
      if (isAncestor(root, i.first)) {
        k = 2;
        colour[i.second] = 2;
      } else {
        from[i.first] = root;
        colour[i.second] = 1;
      }
    } else {
      colour[i.second] = 1;
      dfs(i.first, root);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    edges[i] = {u, v};
    graph[u].push_back({v, i});
  }
  for (int i = 1; i <= n; i++) from[i] = i;
  for (int i = 1; i <= n; i++) {
    if (!seen[i]) dfs(i, i);
  }
  cout << k << '\n';
  for (int i = 0; i < m; i++) cout << colour[i] << ' ';
  cout << '\n';
  return 0;
}