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 = 5005;
vector<pair<int, int> > adj[N];
int cycle[N], mx = 1;
int n, m, res[N];
void dfs(int u) {
  cycle[u] = 1;
  for (int i = 0; i < (int)adj[u].size(); ++i) {
    int v = adj[u][i].second;
    int first = adj[u][i].first;
    if (cycle[v] == 1) {
      res[first] = 2;
      continue;
    } else {
      res[first] = 1;
      if (cycle[v] == 2) continue;
      dfs(v);
    }
  }
  cycle[u] = 2;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(pair<int, int>(i, v));
  }
  for (int i = 1; i <= n; ++i)
    if (cycle[i] == 0) dfs(i);
  mx = *max_element(res + 1, res + m + 1);
  printf("%d\n", mx);
  for (int i = 1; i <= m; ++i) {
    char c = (i == m ? '\n' : ' ');
    printf("%d%c", res[i], c);
  }
  return 0;
}