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, vis[5005], ans[5005], cnt = 1;
vector<pair<int, int>> v[5005];
void dfs(int a) {
  vis[a] = 1;
  for (auto& i : v[a])
    if (vis[i.first] == 0) {
      ans[i.second] = 1;
      dfs(i.first);
    } else if (vis[i.first] == 1) {
      ans[i.second] = 2;
      cnt = 2;
    } else
      ans[i.second] = 1;
  vis[a] = 2;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; i++) {
    int ta, tb;
    scanf("%d%d", &ta, &tb);
    v[ta].emplace_back(tb, i);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i);
  printf("%d\n", cnt);
  for (int i = 0; i < m; i++) printf("%d%c", ans[i], " \n"[i == m - 1]);
}
