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 int;
using pi = pair<int, int>;
const int N = 5100;
int ec[N], ecc, n, m, cm, u, v;
enum class state { not_visited, visiting, visited };
struct node {
  struct edge {
    node* nd;
    int idx;
  };
  vector<edge> adj;
  state s;
  void dfs() {
    s = state::visiting;
    for (auto e : adj) {
      if (ec[e.idx] != 0 || e.nd->s == state::visited) {
        continue;
      }
      if (e.nd->s == state::visiting) {
        ec[e.idx] = -1;
      } else {
        e.nd->dfs();
      }
    }
    s = state::visited;
  }
} a[N];
void dfs() {
  ecc++;
  for (int i = 1; i <= n; i++) {
    a[i].s = state::not_visited;
  }
  for (int i = 1; i <= m; i++) {
    if (ec[i] == -1) {
      ec[i] = 0;
    }
  }
  for (int i = 1; i <= n; i++) {
    if (a[i].s == state::not_visited) {
      a[i].dfs();
    }
  }
  for (int i = 1; i <= m; i++) {
    if (ec[i] == 0) {
      ec[i] = ecc;
      cm++;
    } else if (ec[i] == -1) {
      ec[i] = 0;
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false), cout.tie(nullptr), cin.tie(nullptr);
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    cin >> u >> v;
    a[u].adj.push_back({a + v, i});
  }
  while (cm != m) {
    dfs();
  }
  cout << ecc << '\n';
  for (int i = 1; i <= m; i++) {
    cout << ec[i] << ' ';
  }
}