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;
const long long N = 1e4 + 5;
bool f = false;
vector<pair<long long, long long> > G[N];
long long n, m, u, v, ans[N], vis[N];
void dfs(long long u) {
  vis[u] = 1;
  for (long long i = 0; i < G[u].size(); i++) {
    long long v = G[u][i].first, idx = G[u][i].second;
    if (vis[v] == 0 || vis[v] == 2) {
      ans[idx] = 1;
      if (!vis[v]) {
        dfs(v);
      }
    } else if (vis[v] == 1) {
      f = true;
      ans[idx] = 2;
    }
  }
  vis[u] = 2;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  memset(ans, 0, sizeof(ans));
  memset(vis, 0, sizeof(vis));
  for (long long i = 1; i <= m; i++) {
    cin >> u >> v;
    G[u].push_back(make_pair(v, i));
  }
  for (long long i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i);
    }
  }
  cout << (f ? 2 : 1) << endl;
  for (long long i = 1; i <= m; i++) {
    cout << ans[i] << " ";
  }
  return 0;
}