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;
vector<vector<pair<int, int> > > graph;
int OnStack[5002];
int ans[5002];
bool flag = false;
vector<pair<int, int> > Color2;
void war(int node) {
  OnStack[node] = 1;
  for (int i = 0; i < graph[node].size(); i++) {
    if (OnStack[graph[node][i].first] == 0) {
      ans[graph[node][i].second] = 1;
      war(graph[node][i].first);
    } else if (OnStack[graph[node][i].first] == 1) {
      ans[graph[node][i].second] = 2;
      flag = true;
    } else {
      ans[graph[node][i].second] = 1;
    }
  }
  OnStack[node] = 2;
}
int main() {
  int n, m;
  cin >> n >> m;
  graph.resize(n + 1);
  int a, b;
  for (int i = 1; i <= m; i++) {
    cin >> a >> b;
    graph[a].push_back(pair<int, int>(b, i));
  }
  for (int i = 1; i <= n; i++) {
    if (OnStack[i] == 0) {
      war(i);
    }
  }
  if (flag)
    cout << 2 << endl;
  else
    cout << 1 << endl;
  for (int i = 1; i <= m; i++) cout << ans[i] << " ";
  return 0;
}