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;
using ll = long long;
using pii = pair<int, int>;
const int oo = 1e9 + 4;
const int SZ = 5e3 + 4;
int n, m;
vector<pair<pii, int> > ed;
void read() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int f, t;
    cin >> f >> t;
    ed.push_back({{f - 1, t - 1}, i});
  }
}
vector<int> g[3][SZ];
int us[SZ], CNT;
bool ok;
void dfs(int v, int gr = 1) {
  if (us[v] / 3 == CNT) {
    if (us[v] % 3 == 1) ok = 1;
    return;
  }
  us[v] = CNT * 3 + 1;
  for (auto to : g[gr][v]) dfs(to, gr);
  us[v] = CNT * 3 + 2;
}
int k = 1;
int ans[SZ];
void solve() {
  random_shuffle(ed.begin(), ed.end());
  for (auto t : ed) {
    int f = t.first.first, r = t.first.second, ind = t.second;
    g[1][f].push_back(r);
    ok = 0;
    CNT++;
    for (int i = 0; i < n; i++) dfs(i);
    ans[ind] = 1;
    if (ok) {
      g[1][f].pop_back();
      g[2][f].push_back(r);
      k = 2;
      ans[ind] = 2;
    }
  }
  cout << k << endl;
  for (int i = 0; i < m; i++) cout << ans[i] << ' ';
}
signed main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  read();
  solve();
}