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;
const int N = 1e5 + 100;
int h[N], vis[N], n, m, k, ins[N];
int col[N];
int cnt;
struct edge {
  int to, next;
} e[N << 1];
void add(int u, int v) {
  e[++cnt] = (edge){v, h[u]};
  h[u] = cnt;
}
void dfs(int x) {
  vis[x] = 1;
  ins[x] = 1;
  for (int i = h[x]; i; i = e[i].next)
    if (!vis[e[i].to]) {
      dfs(e[i].to);
    } else if (ins[e[i].to]) {
      col[i] = 1;
    }
  ins[x] = 0;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i);
  int ans = 1;
  for (int i = 1; i <= cnt; i++) ans = max(ans, col[i] + 1);
  printf("%d\n", ans);
  for (int i = 1; i <= cnt; i++) printf("%d ", col[i] + 1);
  puts("");
}
