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 maxn = 5000 + 5;
const int maxm = 100 + 5;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const double pi = acos(-1.0);
int n, m, u, v, flag;
int deg[maxn], col[maxn];
vector<int> g[maxn];
int topo() {
  memset(deg, 0, sizeof deg);
  for (int i = 1; i <= n; i++)
    for (auto it : g[i]) deg[it]++;
  queue<int> q;
  int num = 0;
  for (int i = 1; i <= n; i++)
    if (!deg[i]) q.push(i);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    num++;
    for (auto it : g[u]) {
      if (--deg[it] == 0) q.push(it);
    }
  }
  return num != n;
}
int main() {
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &u, &v);
    g[u].push_back(v);
    col[i] = 1;
    if (topo()) {
      flag = 1;
      g[u].pop_back();
      col[i] = 2;
    }
  }
  if (flag)
    puts("2");
  else
    puts("1");
  for (int i = 1; i <= m; i++) printf("%d ", col[i]);
  return 0;
}