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>
const int N = 5010;
const int M = 5010;
using namespace std;
int vis[N];
int n, m, ans = 1;
int edge[M], lst[N], nxt[M], col[M], t = 0;
void ADD(int x, int y) {
  edge[++t] = y;
  nxt[t] = lst[x];
  lst[x] = t;
  col[t] = 1;
}
void SEARCH(int x) {
  vis[x] = 1;
  for (int r = lst[x]; r; r = nxt[r]) {
    if (vis[edge[r]] == 1) {
      ans = 2;
      col[r] = 2;
      continue;
    }
    if (vis[edge[r]] == -1) continue;
    SEARCH(edge[r]);
  }
  vis[x] = -1;
}
void READ() {
  int u, v;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) scanf("%d%d", &u, &v), ADD(u, v);
}
int main() {
  READ();
  for (int i = 1; i <= n; i++)
    if (!vis[i]) SEARCH(i);
  printf("%d\n", ans);
  for (int i = 1; i <= t; i++) printf("%d ", col[i]);
  return 0;
}
