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;
int head[200010], to[200010], nxt[200010], idx;
int shad[5010][5010], ans[200010], all, n, m;
bool vis[200010], tmp[200010];
void add(int a, int b) { nxt[++idx] = head[a], to[idx] = b, head[a] = idx; }
void dfs(int p, int from) {
  vis[p] = tmp[p] = 1;
  for (int i = head[p]; i; i = nxt[i]) {
    if (!tmp[to[i]]) ans[shad[p][to[i]]] = 1, all++;
    if (!vis[to[i]]) dfs(to[i], p);
  }
  tmp[p] = 0;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1, x, y; i <= m; i++)
    scanf("%d%d", &x, &y), shad[x][y] = i, add(x, y);
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i, 0);
  if (all == m)
    printf("1\n");
  else
    printf("2\n");
  for (int i = 1; i <= m; i++)
    if (!ans[i])
      printf("2 ");
    else
      printf("1 ");
}
