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;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  if (0) return;
  cout << name << " : " << arg1 << '\n';
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  if (0) return;
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
inline void INP();
const long long N = 3e6 + 7, M = 1000000007, inf = INT_MAX;
const long long INF = LLONG_MAX;
long long n, Q, f;
long long arr[N + 2], active[N], vis[N];
vector<vector<pair<long long, long long> > > adj;
void dfs(long long s, long long col) {
  if (vis[s]) {
    if (active[s]) f = 1;
    return;
  }
  vis[s] = 1;
  active[s] = 1;
  for (auto i : adj[s]) {
    if (active[i.first])
      arr[i.second] = 2;
    else
      arr[i.second] = 1;
    dfs(i.first, 1 - col);
  }
  active[s] = 0;
}
void solve() {
  long long n, m;
  cin >> n >> m;
  adj.resize(n);
  for (long long i = 0; i < m; i++) {
    long long u, v;
    cin >> u >> v;
    --u;
    --v;
    adj[u].push_back({v, i});
  }
  for (long long i = 0; i < (long long)n; i++) {
    if (!vis[i]) dfs(i, 0);
  }
  if (!f) {
    cout << 1 << '\n';
    for (long long i = 0; i < m; i++) {
      cout << 1 << " ";
    }
  } else {
    cout << 2 << '\n';
    for (long long i = 0; i < m; i++) {
      cout << arr[i] << " ";
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  INP();
  long long test = 1;
  while (test--) solve();
}
inline void INP() {}