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;
int cnt[5000 + 5];
vector<pair<int, int> > v1[5000 + 5];
vector<pair<int, int> > rev[5000 + 5];
queue<int> kol;
bool col[5000 + 5];
int ans[5000 + 5];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    cnt[a]++;
    v1[a].push_back({b, i});
    rev[b].push_back({a, i});
  }
  for (int i = 1; i <= n; i++)
    if (cnt[i] == 0) kol.push(i);
  int cntDel = 0;
  while (kol.size()) {
    cntDel++;
    int w = kol.front();
    kol.pop();
    for (int i = 0; i < (int)rev[w].size(); i++) {
      int v = rev[w][i].first;
      cnt[v]--;
      if (cnt[v] == 0) kol.push(v);
    }
  }
  if (cntDel == n) {
    cout << 1 << '\n';
    for (int i = 1; i <= m; i++) cout << 1 << ' ';
  } else {
    cout << 2 << '\n';
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < (int)v1[i].size(); j++) {
        int id = v1[i][j].second;
        if (col[id] == 0) {
          col[id] = 1;
          ans[id] = 1;
        }
      }
      for (int j = 0; j < (int)rev[i].size(); j++) {
        int id = rev[i][j].second;
        if (col[id] == 0) {
          col[id] = 1;
          ans[id] = 2;
        }
      }
    }
    for (int i = 1; i <= m; i++) cout << ans[i] << ' ';
  }
  return 0;
}