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 N = 5050;
vector<int> g[N];
int visit[N], finish[N];
bool dfs(int u) {
  if (visit[u]++) return false;
  for (int v : g[u]) {
    if (visit[v] and !finish[v]) return true;
    if (dfs(v)) return true;
    ;
  }
  finish[u]++;
  return false;
}
int main() {
  int n, m;
  cin >> n >> m;
  vector<pair<int, int>> e(m);
  for (auto &e1 : e) cin >> e1.first >> e1.second;
  for (auto &e1 : e) g[e1.first].push_back(e1.second);
  for (int i = 1; i <= n; i++)
    if (dfs(i)) {
      cout << 2 << endl;
      for (int i = 0; i < m; i++)
        cout << (e[i].first < e[i].second ? 1 : 2) << ' ';
      cout << endl;
      return 0;
    }
  cout << 1 << endl;
  for (int i = 0; i < m; i++) cout << 1 << ' ';
  cout << endl;
}
