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;
inline long long read() {
  long long x(0), neg(1);
  char ch(getchar());
  while (!isdigit(ch)) {
    if (ch == '-') neg = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = (x << 1) + (x << 3) + (ch ^ 48);
    ch = getchar();
  }
  return x * neg;
}
long long n, m, cnt, ans = 1, vis[5005], col[5005], head[10005];
struct node {
  long long to, next;
} edge[10005];
inline void add(long long u, long long v) {
  edge[++cnt] = (node){v, head[u]};
  head[u] = cnt;
}
inline void dfs(long long u) {
  vis[u] = 1;
  for (long long i = head[u]; ~i; i = edge[i].next) {
    if (col[i]) continue;
    long long v = edge[i].to;
    if (!vis[v]) dfs(v);
    col[i] = 1;
    if (vis[v] == 1) col[i] = ans = 2;
  }
  vis[u] = 2;
}
signed main() {
  memset(head, -1, sizeof(head));
  long long n = read(), m = read();
  for (long long i = 1; i <= m; ++i) {
    long long u = read(), v = read();
    add(u, v);
  }
  for (long long i = 1; i <= n; ++i) {
    if (!vis[i]) dfs(i);
  }
  printf("%lld\n", ans);
  for (long long i = 1; i <= m; ++i) {
    printf("%lld ", col[i]);
  }
  return 0;
}
