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;
map<pair<int, int>, int> type;
vector<int> g[200100];
int used[200100];
vector<int> tp;
void dfs1(int v) {
  used[v] = 1;
  for (auto to : g[v]) {
    if (!used[to]) dfs1(to);
  }
  tp.push_back(v);
}
void dfs2(int v) {
  used[v] = 1;
  for (auto to : g[v]) {
    if (!used[to]) {
      type[make_pair(v, to)] = 0;
      dfs2(to);
    } else if (used[to] != 1)
      type[make_pair(v, to)] = 0;
  }
  used[v] = 2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > ans;
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    ans.push_back(make_pair(a, b));
    g[a].push_back(b);
    type[make_pair(a, b)] = 1;
  }
  for (int i = 1; i <= n; i++)
    if (!used[i]) dfs1(i);
  reverse(tp.begin(), tp.end());
  for (int i = 1; i <= n; i++) used[i] = 0;
  for (int i = 0; i < tp.size(); i++) {
    if (!used[tp[i]]) {
      dfs2(tp[i]);
    }
  }
  vector<int> print;
  int d = -1;
  bool bad = 1;
  for (int i = 0; i < ans.size(); i++) {
    print.push_back(type[ans[i]]);
    if (d != -1 && print.back() != d)
      bad = 0;
    else if (d == -1)
      d = print.back();
  }
  cout << 2 - bad << endl;
  for (int i = 0; i < print.size(); i++) {
    cout << print[i] + 1 << " ";
  }
  cout << endl;
  return 0;
}