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 N = 5000;
vector<int> g[N];
vector<int> topSort;
bool used[N];
void dfs(int v) {
  used[v] = true;
  for (int u : g[v])
    if (!used[u]) dfs(u);
  topSort.push_back(v);
}
int pos[N];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  vector<pair<int, int> > edges;
  for (int i = 0; i < m; ++i) {
    static int u, v;
    scanf("%d%d", &u, &v);
    --u;
    --v;
    edges.push_back({u, v});
    g[u].push_back(v);
  }
  for (int v = 0; v < n; ++v)
    if (!used[v]) dfs(v);
  reverse(topSort.begin(), topSort.end());
  for (int i = 0; i < n; ++i) pos[topSort[i]] = i;
  int ans = 1;
  for (auto e : edges) {
    if (pos[e.first] > pos[e.second]) {
      ans = 2;
      break;
    }
  }
  printf("%d\n", ans);
  for (auto e : edges) {
    if (pos[e.first] > pos[e.second])
      printf("2 ");
    else
      printf("1 ");
  }
}