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;
int n, m, k = 1;
vector<vector<pair<int, int>>> g;
int visit[5001];
int c[5001];
void dfs(int s) {
  int to, id;
  visit[s] = 1;
  for (auto edge : g[s]) {
    to = edge.first;
    id = edge.second;
    if (visit[to] == 0) {
      dfs(to);
    } else if (visit[to] == 1) {
      c[id] = 2;
      k = 2;
    }
  }
  visit[s] = 2;
}
int main() {
  int i, a, b, j;
  cin >> n >> m;
  g.resize(n + 1);
  for (i = 0; i < m; ++i) {
    cin >> a >> b;
    a--;
    b--;
    g[a].push_back(make_pair(b, i));
  }
  for (i = 0; i < n; ++i) {
    visit[i] = 0;
  }
  for (i = 0; i < m; ++i) {
    c[i] = 1;
  }
  for (i = 0; i < n; ++i) {
    if (visit[i] == 0) {
      dfs(i);
    }
  }
  cout << k << endl;
  for (i = 0; i < m; ++i) {
    cout << c[i] << " ";
  }
  return 0;
}
