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>
#pragma GCC optimize(3)
using namespace std;
int p, m;
int b[5010];
vector<int> v[5010];
vector<pair<int, int> > w;
map<pair<int, int>, int> mpp;
int ans = 1;
inline void dfs(int n) {
  b[n] = 1;
  for (int i = 0; i < v[n].size(); i++) {
    if (b[v[n][i]] == 1) {
      mpp[make_pair(n, v[n][i])] = 2;
      ans = 2;
    }
    if (b[v[n][i]] == 0) {
      mpp[make_pair(n, v[n][i])] = 1;
      dfs(v[n][i]);
    }
    if (b[v[n][i]] == 2) {
      mpp[make_pair(n, v[n][i])] = 1;
    }
  }
  b[n] = 2;
}
int main(int argc, char** argv) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> p >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    w.push_back(make_pair(a, b));
  }
  for (int i = 1; i <= p; i++) {
    if (b[i] == 0) {
      dfs(i);
    }
  }
  cout << ans << endl;
  for (int i = 0; i < m; i++) {
    cout << mpp[w[i]] << " ";
  }
  return 0;
}
