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;
int n, m;
vector<pair<int, int> > niz[5005];
int tip[5005];
int obidjen[5005];
bool ima;
void dfs(int t, int o) {
  obidjen[t] = 1;
  for (auto x : niz[t]) {
    if (obidjen[x.first] == 1) {
      tip[x.second] = 2;
      ima = true;
    } else
      tip[x.second] = 1;
    if (!obidjen[x.first]) dfs(x.first, t);
  }
  obidjen[t] = 2;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int t1, t2;
    scanf("%d %d", &t1, &t2);
    niz[t1].push_back(make_pair(t2, i));
  }
  for (int i = 1; i <= n; i++)
    if (!obidjen[i]) dfs(i, 0);
  if (ima)
    printf("2\n");
  else
    printf("1\n");
  for (int i = 1; i <= m; i++) printf("%d ", tip[i]);
  return 0;
}
