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 
#include <bits/stdc++.h>
using namespace std;
long long spf[101];
long long fac[101];
void sieve() {
  spf[1] = 1;
  for (long long i = 2; i < 101; i++) spf[i] = i;
  for (long long i = 4; i < 101; i += 2) spf[i] = 2;
  for (long long i = 3; i * i < 101; i++) {
    if (spf[i] == i) {
      for (long long j = i * i; j < 101; j += i)
        if (spf[j] == j) spf[j] = i;
    }
  }
}
map<long long, long long> getfactor(long long a) {
  map<long long, long long> m;
  while (a > 1) {
    m[spf[a]]++;
    a /= spf[a];
  }
  return m;
}
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  if (x == 0 || y < 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
long long inverse(long long a, long long p) { return power(a, p - 2, p); }
long long ncr(long long n, long long r, long long p) {
  if (r > n) return 0;
  if (r == 0 || r == n) return 1;
  return (fac[n] * inverse(fac[r], p) % p * inverse(fac[n - r], p) % p) % p;
}
vector<long long> v[5001];
long long ans[5001];
bool check[5001], vis[5001];
map<pair<long long, long long>, long long> m;
void dfs(long long cur) {
  vis[cur] = 1;
  check[cur] = 1;
  for (auto it : v[cur]) {
    if (vis[it] == 0) {
      ans[m[make_pair(cur, it)]] = 1;
      dfs(it);
    } else if (vis[it] == 1 && check[it] == 1) {
      ans[m[make_pair(cur, it)]] = 2;
    } else
      ans[m[make_pair(cur, it)]] = 1;
  }
  check[cur] = 0;
}
void solve() {
  long long n, ma;
  cin >> n >> ma;
  for (long long i = 1; i <= ma; i++) {
    long long x, y;
    cin >> x >> y;
    m[make_pair(x, y)] = i;
    v[x].push_back(y);
  }
  for (long long i = 1; i <= n; i++)
    if (!vis[i]) dfs(i);
  long long maxi = 0;
  for (long long i = 1; i <= ma; i++) maxi = max(ans[i], maxi);
  cout << maxi << '\n';
  for (long long i = 1; i <= ma; i++) cout << ans[i] << " ";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
}
