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>
#pragma GCC optimize("O2")
using namespace std;
const int N = 5e3 + 1;
vector<int> co[N];
vector<int> vis(N, 0);
vector<int> eul;
void dfs(int x) {
  vis[x] = 1;
  for (auto y : co[x])
    if (!vis[y]) dfs(y);
  eul.push_back(x);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<pair<int, int>> ed(m);
  for (auto &x : ed) {
    cin >> x.first >> x.second;
    co[x.first].push_back(x.second);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i);
  reverse(eul.begin(), eul.end());
  vector<int> pos(n + 1);
  for (int i = 0; i < n; i++) pos[eul[i]] = i;
  vector<int> ans(m);
  int used = 1;
  for (int i = 0; i < m; i++) {
    auto x = ed[i];
    if (pos[x.first] < pos[x.second]) {
      ans[i] = 1;
    } else {
      used = 2;
      ans[i] = 2;
    }
  }
  cout << used << "\n";
  for (auto x : ans) cout << x << " ";
}