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>
#pragma GCC optimize("O3")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
long long color[5000] = {}, pp = 1;
long long ans[5000] = {};
map<pair<long long, long long>, long long> ma;
set<long long> c;
long long f = 1;
long long n, m;
vector<long long> v[5000];
void dfs(long long u) {
  color[u] = 1;
  for (auto i : v[u]) {
    if (color[i] == 1) {
      ans[ma[{u, i}]] = 2;
      f = 2;
    }
    if (!color[i]) dfs(i);
  }
  color[u] = 2;
}
int main() {
  cin >> n >> m;
  for (long long i = 0; i < m; i++) {
    long long a, b;
    cin >> a >> b;
    v[--a].push_back(--b);
    ma[{a, b}] = i;
  }
  for (long long i = 0; i < n; i++)
    if (!color[i]) {
      dfs(i);
    }
  cout << f << endl;
  long long yy = 1;
  for (long long i = 0; i < m; i++) cout << max(ans[i], yy) << " ";
}
