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 iinf = 1e9 + 7;
const long long int inf = 1e9 + 7;
const long long int mod = 998244353;
const long long int maxn = 2e5 + 5;
long long int sqr(long long int x) { return x * x; }
long long int modd(long long int a) {
  if (a > 0) return a % inf;
  while (a < 0) {
    a += inf;
  }
  return a;
}
long long int modexpo(long long int a, long long int b) {
  a %= inf;
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = res * a % inf;
    a = a * a % inf;
    b >>= 1;
  }
  return res;
}
const int man = 5e3 + 5;
vector<pair<int, int>> adj[man];
vector<int> col(man, 0);
vector<int> res(man, 0);
bool cycle;
void dfs(int s) {
  col[s] = 1;
  for (auto it : adj[s]) {
    int to = it.first, id = it.second;
    if (col[to] == 0) {
      res[id] = 1;
      dfs(to);
    } else if (col[to] == 2) {
      res[id] = 1;
    } else {
      cycle = true;
      res[id] = 2;
    }
  }
  col[s] = 2;
}
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back({v, i});
  }
  cycle = false;
  for (int i = 1; i <= n; i++) {
    if (col[i] == 0) {
      dfs(i);
    }
  }
  if (cycle) {
    cout << "2" << endl;
  } else {
    cout << "1" << endl;
  }
  for (int i = 0; i < m; i++) {
    cout << res[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
}