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;
long long n, m;
vector<long long int> vis(5002);
vector<long long int> rec(5002);
vector<long long int> adj[5002];
map<pair<long long int, long long int>, long long> ans;
bool tr = false;
void dfs(long long x) {
  vis[x] = 1;
  rec[x] = 1;
  for (auto i : adj[x]) {
    if (rec[i]) {
      ans[{x, i}] = 2;
      tr = true;
    } else if (!vis[i]) {
      ans[{x, i}] = 1;
      dfs(i);
    }
  }
  rec[x] = 0;
}
int main() {
  cin >> n >> m;
  long long x, y;
  vector<pair<long long int, long long int> > sol;
  for (long long int i = 0; i < m; i++) {
    cin >> x >> y;
    sol.push_back({x, y});
    adj[x].push_back(y);
  }
  priority_queue<pair<long long int, long long int> > pq;
  for (long long int i = 0; i < n; i++) {
    pq.push({adj[i].size(), i});
  }
  while (!pq.empty()) {
    pair<long long int, long long int> cur = pq.top();
    pq.pop();
    if (vis[cur.second]) continue;
    dfs(cur.second);
  }
  if (tr)
    cout << 2 << "\n";
  else
    cout << 1 << "\n";
  for (long long int i = 0; i < m; i++) {
    if (ans[sol[i]] == 0)
      cout << 1 << " ";
    else
      cout << ans[sol[i]] << " ";
  }
}