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 MAXN = 6000;
int availability[MAXN][MAXN];
bool used[MAXN];
vector<int> g[MAXN];
int for_dfs[MAXN];
vector<pair<int, int>> for_del;
void dfs(int v) {
  for_dfs[v] = 1;
  for (int i = 0; i < (int)g[v].size();) {
    if (for_dfs[g[v][i]] == 1)
      i++;
    else {
      if (for_dfs[g[v][i]] == 0) dfs(g[v][i]);
      for_del.emplace_back(v, g[v][i]);
      g[v].erase(g[v].begin() + i);
    }
  }
  for_dfs[v] = 2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  vector<pair<int, int>> edges;
  for (int i = 0; i < m; i++) {
    int v, u;
    cin >> v >> u;
    g[v].push_back(u);
    edges.emplace_back(v, u);
  }
  int c = 1;
  int added = 0;
  vector<int> ans(m);
  while (added < m) {
    memset(for_dfs, 0, sizeof(for_dfs));
    for (int i = 1; i <= n; i++) {
      if (!for_dfs[i]) {
        dfs(i);
      }
    }
    for (auto edge : for_del) {
      ans[find(edges.begin(), edges.end(), edge) - edges.begin()] = c;
    }
    added += for_del.size();
    c++;
    for_del.clear();
  }
  cout << *max_element(ans.begin(), ans.end()) << endl;
  assert(*max_element(ans.begin(), ans.end()) <= m);
  for (int i = 0; i < m; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}
