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 long long MAX = 1e6 + 8;
const long long MOD = 1e9 + 7;
const long long sz = 1e5 + 5;
void f_io() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
}
vector<long long> col(MAX, 0);
vector<pair<long long, long long>> g[MAX];
vector<long long> vis(MAX, 0);
long long ret = 1;
void dfs(long long node) {
  vis[node] = true;
  for (auto nei : g[node]) {
    if (!vis[nei.first])
      dfs(nei.first);
    else if (vis[nei.first] == 1) {
      ret = 2;
      col[nei.second] = 2;
    }
  }
  vis[node] = 2;
}
int32_t main() {
  f_io();
  long long tc = 1;
  while (tc--) {
    long long n, m;
    cin >> n >> m;
    for (long long u, v, i = 0; i < m; i++) {
      cin >> u >> v;
      g[u].push_back({v, i + 1});
      col[i + 1] = 1;
    }
    for (long long i = 1; i < n + 1; i++) {
      if (!vis[i]) dfs(i);
    }
    cout << ret << "\n";
    for (long long i = 1; i < m + 1; i++) cout << col[i] << " ";
  }
}