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;
vector<pair<long long, long long> > ad[5005];
int vis[5005];
long long k = 0;
int c[5005];
bool f = false;
void dfs(long long i) {
  vis[i] = 1;
  for (auto u : ad[i]) {
    long long r = u.first, j = u.second;
    if (vis[r] == 0) {
      c[j] = 1;
      dfs(r);
    } else if (vis[r] == 1) {
      c[j] = 2;
      f = true;
    } else {
      c[j] = 1;
    }
  }
  vis[i] = 2;
  return;
}
int main() {
  sizeof(vis, 0, sizeof(vis));
  long long n, m;
  cin >> n >> m;
  for (long long i = 1; i < m + 1; i = i + 1) {
    long long l, r;
    cin >> l >> r;
    ad[l].push_back(make_pair(r, i));
  }
  for (long long i = 1; i < n + 1; i = i + 1) {
    if (vis[i] == 0) {
      dfs(i);
    }
  }
  if (!f) {
    cout << 1 << endl;
    for (long long i = 0; i < m; i = i + 1) cout << 1 << " ";
    return 0;
  } else {
    cout << 2 << endl;
    for (long long i = 1; i < m + 1; i = i + 1) cout << c[i] << " ";
  }
}
