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>
#pragma GCC optimze("Ofast")
using namespace std;
const long long infl = 0x3f3f3f3f3f3f3f3fLL;
const int infi = 0x3f3f3f3f;
const int mod = 1000000007;
int powm(int a, int b) {
  int res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
vector<pair<int, int> > edge[5001];
int vis[5001];
int ans[5001];
bool cycle;
void dfs(int v) {
  vis[v] = 1;
  for (pair<int, int> p : edge[v]) {
    int to = p.first, id = p.second;
    if (vis[to] == 0) {
      dfs(to);
      ans[id] = 1;
    } else if (vis[to] == 1) {
      ans[id] = 2;
      cycle = true;
    } else {
      ans[id] = 1;
    }
  }
  vis[v] = 2;
}
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    edge[x].push_back(pair<int, int>(y, i));
  }
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) dfs(i);
  }
  cout << (cycle ? 2 : 1) << '\n';
  for (int i = 1; i <= m; i++) cout << ans[i] << " ";
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(12);
  int t = 1;
  while (t--) solve();
  return 0;
}