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;
using ll = long long;
using vi = vector<int>;
using pi = pair<int, int>;
const int mod = 1e9 + 7;
const ll inf = ll(1e18);
const double PI = acos(-1), eps = 1e-9;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int mxn = 5e3 + 5, mxa = 1e3, mxk = 5;
const int d4i[] = {-1, 0, 1, 0}, d4j[] = {0, 1, 0, -1};
int n, m, vis[mxn];
vi adj[mxn];
bool ok = 1;
void dfs(int v) {
  vis[v] = 1;
  for (const int x : adj[v]) {
    if (vis[x] == 1)
      ok = 0;
    else if (!vis[x])
      dfs(x);
  }
  vis[v] = 2;
}
void solve() {
  cin >> n >> m;
  vector<pi> ve(m);
  for (int i = 0; i < m; ++i) {
    cin >> ve[i].first >> ve[i].second;
    --ve[i].first, --ve[i].second;
    adj[ve[i].first].push_back(ve[i].second);
  }
  for (int i = 0; i < n; ++i) {
    if (!vis[i]) dfs(i);
  }
  if (ok) {
    cout << "1\n";
    for (int i = 0; i < m; ++i) cout << "1 ";
    cout << '\n';
    return;
  }
  cout << "2\n";
  for (int i = 0; i < m; ++i) {
    cout << (ve[i].first > ve[i].second ? 1 : 2) << ' ';
  }
  cout << '\n';
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int T = 1;
  for (int tt = 1; tt <= T; ++tt) {
    solve();
  }
}