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 main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<int> ans(m);
  vector<vector<int> > edges(n);
  vector<int> deg(n, 0);
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    deg[v]++;
    ans[i] = (u < v);
    edges[u].push_back(v);
  }
  vector<int> bfs;
  for (int i = 0; i < n; i++) {
    if (deg[i] == 0) bfs.push_back(i);
  }
  int s = 0;
  while (s < (int)bfs.size()) {
    int v = bfs[s];
    s += 1;
    for (int w : edges[v]) {
      deg[w]--;
      if (deg[w] == 0) bfs.push_back(w);
    }
  }
  if (n == (int)bfs.size()) {
    cout << 1 << '\n';
    for (int x : ans) cout << 1 << ' ';
    cout << '\n';
  } else {
    cout << 2 << '\n';
    for (int x : ans) cout << 1 + x << ' ';
    cout << '\n';
  }
}
