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 
#include <bits/stdc++.h>
using namespace std;
const long double PI = acos(-1.0);
vector<long long> adjlist[5005];
long long pos[5005];
bool vis[5005];
vector<long long> topsort;
void dfs(long long node) {
  vis[node] = 1;
  for (long long i : adjlist[node]) {
    if (!vis[i]) dfs(i);
  }
  pos[node] = topsort.size();
  topsort.push_back(node);
}
void solve() {
  long long n, m;
  cin >> n >> m;
  vector<pair<long long, long long> > edges;
  for (long long i = 0; i < m; ++i) {
    long long u, v;
    cin >> u >> v;
    adjlist[u].push_back(v);
    edges.push_back(make_pair(u, v));
  }
  for (long long i = 1; i < n + 1; ++i) {
    if (!vis[i]) dfs(i);
  }
  vector<long long> colors;
  long long cycle = 0;
  for (auto i : edges) {
    if (pos[i.first] < pos[i.second]) {
      cycle = 1;
      colors.push_back(2);
    } else
      colors.push_back(1);
  }
  if (cycle)
    cout << 2 << '\n';
  else
    cout << 1 << '\n';
  for (long long i : colors) cout << i << ' ';
  cout << '\n';
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}
