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>
namespace IO {
inline char gc() {
  static char s[1 << 20 | 1] = {0}, *p1 = s, *p2 = s;
  return (p1 == p2) && (p2 = (p1 = s) + fread(s, 1, 1 << 20, stdin), p1 == p2)
             ? EOF
             : *(p1++);
}
inline long long read() {
  long long ret = 0;
  bool flag = 0;
  char c = gc();
  while ((c < '0') | (c > '9')) flag ^= !(c ^ '-'), c = gc();
  while ((c >= '0') & (c <= '9'))
    ret = (ret << 1) + (ret << 3) + (c ^ '0'), c = gc();
  return flag ? -ret : ret;
}
char OutputAns[1 << 20 | 1], *OutputCur = OutputAns;
inline void output() {
  OutputCur -= fwrite(OutputAns, 1, OutputCur - OutputAns, stdout);
}
inline void print(long long ans) {
  char s[20] = {0};
  if (OutputCur - OutputAns + sprintf(s, "I64d", ans) >> 20) output();
  OutputCur += sprintf(OutputCur, "%I64d", ans);
}
inline void printc(char c) {
  if (OutputCur - OutputAns + 1 >> 20) output();
  *(OutputCur++) = c;
}
}  // namespace IO
using IO::output;
using IO::print;
using IO::printc;
using IO::read;
using namespace std;
const int M = 5e3 + 11;
int Nxt[M], To[M], Fir[M], Id[M], Is[M], Ans[M];
bool cyc;
int n, m, tot;
inline void Find(int u) {
  Is[u] = 1;
  for (int i = Fir[u]; i; i = Nxt[i]) {
    int v = To[i], id = Id[i];
    if (!Is[v]) {
      Ans[id] = 1;
      Find(v);
    } else if (Is[v] == 2)
      Ans[id] = 1;
    else
      Ans[id] = 2, cyc = 1;
  }
  Is[u] = 2;
}
inline void AddEdge(int u, int v, int id) {
  Nxt[++tot] = Fir[u], To[tot] = v, Id[tot] = id, Fir[u] = tot;
}
int main(void) {
  n = read(), m = read();
  for (int i = 1; i <= m; ++i) {
    int u = read(), v = read();
    AddEdge(u, v, i);
  }
  for (int i = 1; i <= n; ++i)
    if (!Is[i]) Find(i);
  print(cyc + 1), printc('\n');
  for (int i = 1; i <= m; ++i) print(Ans[i]), printc(i == m ? '\n' : ' ');
  output();
  return 0;
}