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 int mod = 1e9 + 7;
const long long int INF = 9e18 + 2e17;
const int inf = 2e9;
const double eps = 1e-10;
struct edge {
  int id, v;
  edge() {}
  edge(int id, int v) : id(id), v(v) {}
};
int cycle = 0;
void dfs(int u, std::vector<std::vector<edge> >& g, vector<int>& res,
         vector<int>& col) {
  col[u] = 1;
  for (auto& e : g[u]) {
    if (col[e.v] == 0) {
      res[e.id] = 1;
      dfs(e.v, g, res, col);
    } else if (col[e.v] == 1) {
      res[e.id] = 2;
      cycle = 1;
    } else {
      res[e.id] = 1;
    }
  }
  col[u] = 2;
}
int main(int argc, char const* argv[]) {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  std::vector<std::vector<edge> > g(n + 1, std::vector<edge>());
  vector<int> res(m);
  vector<int> col(n + 1);
  for (int i = (int)0; i < (int)m; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(edge(i, v));
  }
  for (int i = (int)1; i < (int)n + 1; i++) {
    if (col[i] == 0) dfs(i, g, res, col);
  }
  if (cycle)
    cout << 2 << endl;
  else {
    cout << 1 << endl;
  }
  for (int i = (int)0; i < (int)m; i++) cout << res[i] << " ";
  cout << endl;
  return 0;
}
