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 N = 5005;
pair<int, int> edge[N];
vector<int> g[N];
vector<int> tsort;
int pos[N], col[N];
bool vis[N];
void dfs(int v) {
  if (vis[v]) return;
  vis[v] = true;
  for (auto u : g[v]) {
    dfs(u);
  }
  tsort.push_back(v);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << setprecision(32);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    edge[i] = {u, v};
    g[u].push_back(v);
  }
  memset(vis, false, sizeof(vis));
  for (int i = 1; i <= n; i++) {
    dfs(i);
  }
  reverse(tsort.begin(), tsort.end());
  for (int i = 0; i < n; i++) {
    pos[tsort[i]] = i;
  }
  bool flg = true;
  for (int i = 1; i <= m; i++) {
    int u = edge[i].first;
    int v = edge[i].second;
    if (pos[u] > pos[v]) {
      flg = false;
      col[i] = 2;
    } else {
      col[i] = 1;
    }
  }
  cout << (flg ? 1 : 2) << '\n';
  for (int i = 1; i <= m; i++) {
    cout << col[i] << " ";
  }
  return 0;
}