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 ru[5005];
vector<int> g[5005];
int lu[5005][2];
int main() {
  int i, j, n, m, x, y;
  int t;
  cin >> n >> m;
  for (i = 1; i <= m; i++) {
    cin >> x >> y;
    lu[i][0] = x;
    lu[i][1] = y;
    ru[y]++;
    g[x].push_back(y);
  }
  queue<int> q;
  int sum = 0;
  for (i = 1; i <= n; i++) {
    if (ru[i] == 0) {
      sum++;
      q.push(i);
    }
  }
  while (!q.empty()) {
    int pp = q.front();
    q.pop();
    for (i = 0; i < g[pp].size(); i++) {
      ru[g[pp][i]]--;
      if (ru[g[pp][i]] == 0) {
        q.push(g[pp][i]);
        sum++;
      }
    }
  }
  if (sum == n) {
    cout << "1" << endl;
    for (i = 1; i <= m; i++)
      cout << "1"
           << " ";
  } else {
    cout << 2 << endl;
    for (i = 1; i <= m; i++) {
      if (lu[i][1] < lu[i][0])
        cout << "1"
             << " ";
      else
        cout << "2"
             << " ";
    }
  }
  return 0;
}
