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>
const int N = 1e6 + 10;
const int mod = 998244353;
using namespace std;
vector<pair<int, int> > v[5500];
int ans[5050], vis[5050], k = 1;
void dfs(int x, int co) {
  vis[x] = 1;
  for (auto e : v[x]) {
    if (vis[e.first] == 1)
      ans[e.second] = co + 1;
    else if (vis[e.first] == 0) {
      ans[e.second] = co;
      dfs(e.first, co);
    } else
      ans[e.second] = k;
  }
  vis[x] = 2;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back({y, i});
  }
  for (int i = 1; i <= n; ++i)
    if (vis[i] == 0) k = 1, dfs(i, 1);
  for (int i = 1; i <= m; ++i) k = max(k, ans[i]);
  printf("%d\n", k);
  for (int i = 1; i <= m; ++i) printf("%d%c", ans[i], " \n"[i == m]);
  return 0;
}