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;
int flag, used[1111111], used1[1111111];
vector<pair<int, int>> g[1111111];
void cycle(int v) {
  used1[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i].first;
    if (used[g[v][i].second] == 1) {
      continue;
    }
    if (used1[to] == 1) {
      used[g[v][i].second] = 1;
      flag = 1;
    }
    if (used1[to] == 0) {
      cycle(to);
    }
  }
  used1[v] = 2;
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int v, u;
    cin >> v >> u;
    g[v].push_back({u, i});
  }
  for (int i = 1; i <= n; i++) {
    if (!used1[i]) {
      cycle(i);
    }
  }
  if (!flag) {
    cout << 1 << endl;
    for (int i = 1; i <= m; i++) {
      cout << 1 << ' ';
    }
    return 0;
  }
  while (flag) {
    flag = 0;
    for (int i = 1; i <= n; i++) {
      if (!used1[i]) {
        cycle(i);
      }
    }
  }
  cout << 2 << endl;
  for (int i = 1; i <= m; i++) {
    cout << used[i] + 1 << ' ';
  }
}
