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 int INF = 0x3f3f3f3f;
const long long LINF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-8;
const int MOD = 1000000007;
const int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
vector<pair<int, int> > graph[5000];
bool visited[5000] = {}, calculated[5000] = {};
int ans[5000];
void dfs(int ver) {
  for (auto pr : graph[ver]) {
    if (visited[pr.first]) {
      ans[pr.second] = (calculated[pr.first] ? 1 : 2);
    } else {
      visited[pr.first] = true;
      ans[pr.second] = 1;
      dfs(pr.first);
    }
  }
  calculated[ver] = true;
}
int main() {
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  for (int i = (0); i < (m); ++i) {
    int u, v;
    cin >> u >> v;
    --u;
    --v;
    graph[u].emplace_back(v, i);
  }
  for (int i = (0); i < (n); ++i) {
    if (!visited[i]) {
      visited[i] = true;
      dfs(i);
    }
  }
  cout << (count(ans, ans + m, 2) > 0 ? 2 : 1) << '\n';
  for (int i = (0); i < (m); ++i) cout << ans[i] << (i + 1 == m ? '\n' : ' ');
  return 0;
}