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;
vector<pair<int, int>> edges[5555];
bool done[5555];
bool dfs(int curr, int target, int c) {
  if (curr == target) return true;
  if (done[curr]) return false;
  done[curr] = true;
  for (pair<int, int> vc : edges[curr]) {
    if (vc.second == c && dfs(vc.first, target, c)) return true;
  }
  return false;
}
int ans[5555];
int main() {
  int n, m;
  cin >> n >> m;
  int maxc = 1;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    for (int c = 1;; c++) {
      maxc = max(maxc, c);
      for (int j = 1; j <= n; j++) done[j] = false;
      if (!dfs(v, u, c)) {
        ans[i] = c;
        edges[u].push_back({v, c});
        break;
      }
    }
  }
  cout << maxc << endl;
  for (int i = 1; i <= m; i++) {
    printf("%d ", ans[i]);
  }
  cout << endl;
}
