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;
const long long mod = 998244353;
const int maxn = 5e3 + 5;
vector<pair<int, int>> g[maxn];
int vis[maxn], ans[maxn];
void dfs(int x) {
  vis[x] = 1;
  for (int i = (0); i < (g[x].size()); i++) {
    int v = g[x][i].first;
    if (!vis[v]) {
      ans[g[x][i].second] = 1;
      dfs(v);
    } else if (vis[v] == 1) {
      ans[g[x][i].second] = 2;
    } else
      ans[g[x][i].second] = 1;
  }
  vis[x] = 2;
}
int main() {
  int n, m;
  scanf("%d", &n);
  scanf("%d", &m);
  for (int i = (1); i <= (m); i++) {
    int u, v;
    scanf("%d", &u);
    scanf("%d", &v);
    g[u].push_back(make_pair(v, i));
  }
  for (int i = (1); i <= (n); i++)
    if (!vis[i]) dfs(i);
  int s = 1;
  for (int i = (1); i <= (m); i++)
    if (ans[i] == 2) {
      s = 2;
      break;
    }
  printf("%d", s);
  puts("");
  for (int i = (1); i <= (m); i++) printf("%d ", ans[i]);
}
