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;
const int P = 1e9 + 7;
inline int add(int a, int b) {
  if ((a += b) >= P) a -= P;
  return a;
}
inline int sub(int a, int b) {
  if ((a -= b) < 0) a += P;
  return a;
}
inline int mul(int a, int b) { return 1ll * a * b % P; }
inline int kpow(int a, int b) {
  int r = 1;
  for (; b; b >>= 1, a = mul(a, a)) {
    if (b & 1) r = mul(r, a);
  }
  return r;
}
const int N = 5050;
int n, m;
vector<int> G[N];
int in[N];
int u[N], v[N];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = (0); i < (m); i++) {
    scanf("%d%d", u + i, v + i);
    u[i]--, v[i]--;
    G[u[i]].push_back(v[i]);
    in[v[i]]++;
  }
  vector<int> q;
  for (int i = (0); i < (n); i++)
    if (in[i] == 0) q.push_back(i);
  for (int f = (0); f < ((int)q.size()); f++) {
    for (int &num : G[q[f]]) {
      in[num]--;
      if (!in[num]) q.push_back(num);
    }
  }
  if ((int)q.size() == n) {
    printf("1\n");
    for (int i = (0); i < (m); i++) printf("1 ");
  } else {
    printf("2\n");
    for (int i = (0); i < (m); i++)
      if (u[i] > v[i])
        printf("%d ", 1);
      else
        printf("%d ", 2);
  }
  return 0;
}