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[5010], to[5010 << 1], nxt[5010 << 1], tot, sum;
inline void add(int x, int y) {
  to[++tot] = y;
  nxt[tot] = head[x];
  head[x] = tot;
}
char *p1, *p2, buf[100000];
int rd() {
  int x = 0;
  char c =
      (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
           ? EOF
           : *p1++);
  while (c < 48) {
    c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++);
  }
  while (c > 47) {
    x = (((x << 2) + x) << 1) + (c ^ 48),
    c = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 100000, stdin), p1 == p2)
             ? EOF
             : *p1++);
  }
  return x;
}
bool v1[5010], v2[5010];
int ar[5010][5010], ans[5010];
void dfs(int p) {
  v1[p] = v2[p] = 1;
  for (int i = head[p]; i; i = nxt[i]) {
    if (!v2[to[i]]) {
      ans[ar[p][to[i]]] = 1;
      sum++;
    }
    if (!v1[to[i]]) {
      dfs(to[i]);
    }
  }
  v2[p] = false;
}
int main() {
  int n = rd(), m = rd();
  for (int i = 1; i <= m; i++) {
    int x = rd(), y = rd();
    ar[x][y] = i;
    add(x, y);
  }
  for (int i = 1; i <= n; i++) {
    if (!v1[i]) {
      dfs(i);
    }
  }
  if (sum == m) {
    puts("1");
  } else {
    puts("2");
  }
  for (int i = 1; i <= m; i++) {
    if (!ans[i]) {
      cout << "2 ";
    } else {
      cout << "1 ";
    }
  }
}
