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 = 2e5;
vector<int> g[N], used, g1[N];
int gl[N];
bool ok = 1;
void dfs(int v, int glb) {
  used[v] = 1;
  gl[v] = glb;
  for (int to : g[v])
    if (!used[to]) dfs(to, glb + 1);
}
void dfs1(int v, int root) {
  used[v] = 1;
  for (int to : g1[v])
    if (to == root)
      ok = 0;
    else if (!used[to])
      dfs1(to, root);
}
int32_t main() {
  ios_base::sync_with_stdio(NULL);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  map<pair<int, int>, int> q;
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    --a;
    --b;
    q[make_pair(a, b)] = i;
    g[a].push_back(b);
    g[b].push_back(a);
    g1[a].push_back(b);
  }
  for (int i = 0; i < n; ++i) {
    used.assign(n, 0);
    dfs1(i, i);
  }
  used.assign(n, 0);
  for (int i = 0; i < n; ++i)
    if (!used[i]) dfs(i, 0);
  if (ok) {
    cout << "1\n";
    for (int i = 0; i < m; ++i) cout << "1 ";
    return 0;
  }
  vector<int> res(m);
  for (auto x : q) {
    int a = x.first.first;
    int b = x.first.second;
    if (gl[a] > gl[b])
      res[x.second] = 1;
    else
      res[x.second] = 2;
  }
  cout << "2\n";
  for (int i = 0; i < m; ++i) cout << res[i] << " ";
}