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 N = 10000;
int res[N], fix[N];
bool cycle;
vector<pair<int, int> > edges[N];
void dfs(int x, bool p) {
  fix[x] = 1;
  for (int i = 0; i < edges[x].size(); i++) {
    int u = edges[x][i].first;
    if (fix[u] == 1)
      cycle = true, res[edges[x][i].second] = (p ^ 1) + 1;
    else
      res[edges[x][i].second] = p + 1;
    if (!fix[u]) dfs(u, p);
  }
  fix[x] = 2;
}
int main() {
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    edges[a].push_back({b, i});
  }
  for (int i = 1; i <= n; i++)
    if (!fix[i]) dfs(i, 0);
  cout << (cycle ? 2 : 1) << "\n";
  for (int i = 1; i <= m; i++) cout << res[i] << " \n"[i == m];
}