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;
vector<string> split(const string& s, char c) {
  vector<string> v;
  stringstream second(s);
  string x;
  while (getline(second, x, c)) v.emplace_back(x);
  return move(v);
}
template <typename T, typename... Args>
inline string arrStr(T arr, int n) {
  stringstream s;
  s << "[";
  for (int i = 0; i < n - 1; i++) s << arr[i] << ",";
  s << arr[n - 1] << "]";
  return s.str();
}
inline void __evars_begin(int line) { cerr << "#" << line << ": "; }
template <typename T>
inline void __evars_out_var(vector<T> val) {
  cerr << arrStr(val, val.size());
}
template <typename T>
inline void __evars_out_var(T* val) {
  cerr << arrStr(val, 10);
}
template <typename T>
inline void __evars_out_var(T val) {
  cerr << val;
}
inline void __evars(vector<string>::iterator it) { cerr << '\n'; }
template <typename T, typename... Args>
inline void __evars(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << "=";
  __evars_out_var(a);
  cerr << "; ";
  __evars(++it, args...);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int dx[8] = {1, -1, 0, 0, 1, 1, -1, -1};
int dy[8] = {0, 0, -1, 1, 2, -2, 2, -2};
vector<long long> vis(200010, 0), dis(200010, 0);
vector<long long> adj[200010];
int p[200010][20];
int myrandom(int i) { return std::rand() % i; }
const int MAX = 200010;
const unsigned long long inf = 1000000007;
set<pair<int, int>> st;
map<pair<int, int>, int> ans;
bool ok = false;
void dfs(int a, int p) {
  vis[a] = 1;
  for (auto i : adj[a]) {
    if (!vis[i]) {
      ans[{a, i}] = 1;
      dfs(i, a);
    } else if (vis[i] == 1) {
      ok = true;
      ans[{a, i}] = 2;
    } else if (vis[i] == 2) {
      ans[{a, i}] = 1;
    }
  }
  vis[a] = 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, m;
  cin >> n >> m;
  vector<pair<int, int>> edg;
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    edg.push_back({a, b});
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i, -1);
  }
  cout << ok + 1 << '\n';
  for (auto i : edg) {
    cout << ans[i] << " ";
  }
  cout << '\n';
}