Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct Edge {
  int v, c, nxt;
} e[200010];
int tot;
int first[100010];
void build(int u, int v, int c) {
  e[++tot] = (Edge){v, c, first[u]};
  first[u] = tot;
  return;
}
int col[100010], flag;
bool ins[100010], vis[100010];
void dfs(int u) {
  vis[u] = true;
  ins[u] = true;
  for (int i = first[u]; i; i = e[i].nxt)
    if (!vis[e[i].v]) {
      col[e[i].c] = 1;
      dfs(e[i].v);
    } else if (ins[e[i].v])
      col[e[i].c] = 2, flag = 1;
    else
      col[e[i].c] = 1;
  ins[u] = false;
  return;
}
int n, m;
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    build(u, v, i);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i);
  printf("%d\n", 1 + flag);
  for (int i = 1; i <= m; i++) printf("%d ", col[i]);
  printf("\n");
  return 0;
}