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;
long long mult(long long a, long long b, long long p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long add(long long a, long long b, long long p = 1000000007) {
  return (a % p + b % p) % p;
}
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1LL) {
    if (k & 1LL) r = mult(r, n, p);
    n = mult(n, n, p);
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
vector<int> v[100001];
vector<pair<int, int> > edges;
int col[100001];
bool cyc;
bool cycle(int s) {
  col[s] = 1;
  for (int i : v[s]) {
    if (col[i] == 1) {
      return true;
    } else if (col[i] == 0 && cycle(i)) {
      return true;
    }
  }
  col[s] = 2;
  return false;
}
void solve() {
  int n, m;
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    edges.push_back(make_pair(a, b));
  }
  for (int i = 1; i < n + 1; i++) {
    if (!col[i] && cycle(i)) {
      cyc = true;
      break;
    }
  }
  if (cyc) {
    cout << 2 << '\n';
    for (int i = 0; i < m; i++) {
      if (edges[i].first > edges[i].second)
        cout << 1 << " ";
      else
        cout << 2 << " ";
    }
    cout << '\n';
  } else {
    cout << 1 << '\n';
    for (int i = 0; i < m; i++) cout << 1 << " ";
    cout << '\n';
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}