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 int MOD7 = 1000000000 + 7;
const int MOD9 = 1000000000 + 9;
vector<pair<int, int> > g[5000];
int status[5000] = {};
int ans[5000] = {};
void dfs(int x) {
  status[x] = 1;
  for (auto e : g[x]) {
    if (status[e.first] == 1) {
      ans[e.second] = 2;
    } else {
      ans[e.second] = 1;
    }
    if (!status[e.first]) {
      dfs(e.first);
    }
  }
  status[x] = 2;
}
int main() {
  int n = ({
    int x;
    scanf("%d", &x);
    x;
  }),
      m = ({
        int x;
        scanf("%d", &x);
        x;
      });
  for (int i = (0); i < (m); i += (1)) {
    int x = ({
      int x;
      scanf("%d", &x);
      x;
    }),
        y = ({
          int x;
          scanf("%d", &x);
          x;
        });
    x--;
    y--;
    g[x].push_back({y, i});
  }
  for (int i = (0); i < (n); i += (1))
    if (!status[i]) dfs(i);
  cout << *max_element(ans, ans + m) << "\n";
  for (int i = (0); i < (m); i += (1)) cout << ans[i] << " ";
  puts("");
}
