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;
struct Edge {
  int to, id;
};
vector<Edge> G[5005];
int col[5005];
bool seen[5005];
bool act[5005];
int k = 0;
void dfs(int at) {
  if (seen[at]) return;
  seen[at] = true;
  act[at] = true;
  for (Edge e : G[at]) {
    if (act[e.to])
      col[e.id] = 2;
    else {
      col[e.id] = 1;
      if (!seen[e.to]) dfs(e.to);
    }
    k = max(k, col[e.id]);
  }
  act[at] = false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int N, M;
  cin >> N >> M;
  for (int i = 0; i < M; i++) {
    int a, b;
    cin >> a >> b;
    G[a - 1].push_back({b - 1, i});
  }
  for (int i = 0; i < N; i++) {
    if (!seen[i]) dfs(i);
  }
  printf("%d\n", k);
  for (int i = 0; i < M; i++) printf("%d ", col[i]);
  printf("\n");
}
