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;
long long int col[5010];
long long int res[5010];
map<pair<long long int, long long int>, long long int> mp;
vector<vector<long long int> > v(5010);
bool tr;
void dfs(long long int x) {
  col[x] = 1;
  for (int i = 0; i < v[x].size(); i++) {
    if (col[v[x][i]] == 0) {
      dfs(v[x][i]);
      res[mp[make_pair(x, v[x][i])]] = 1;
    } else if (col[v[x][i]] == 2) {
      res[mp[make_pair(x, v[x][i])]] = 1;
    } else {
      res[mp[make_pair(x, v[x][i])]] = 2;
      tr = true;
    }
  }
  col[x] = 2;
}
int main() {
  long long int n, m, x, y, i;
  tr = false;
  cin >> n >> m;
  for (i = 0; i < m; i++) {
    cin >> x >> y;
    v[x].push_back(y);
    mp[make_pair(x, y)] = i;
  }
  for (i = 1; i <= n; i++) {
    if (col[i] == 0) dfs(i);
  }
  cout << (tr ? 2 : 1) << endl;
  for (i = 0; i < m; i++) cout << res[i] << " ";
  cout << endl;
}
