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;
template <class T>
inline bool uin(T& a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
inline bool uax(T& a, T b) {
  return a < b ? (a = b, true) : false;
}
const int nax = 5123;
pair<int, int> edge[nax];
vector<int> g[nax];
int col[nax];
int N, M;
bool cycle_found = false;
int dfs_num[nax];
void dfs(int v) {
  dfs_num[v] = 1;
  for (int i : g[v]) {
    int ch = edge[i].first ^ edge[i].second ^ v;
    if (dfs_num[ch] == 2) {
      continue;
    } else if (dfs_num[ch] == 1) {
      col[i] = 1;
      cycle_found = true;
      continue;
    }
    dfs(ch);
  }
  dfs_num[v] = 2;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> N >> M;
  for (int u, v, i = 0; i < M; ++i) {
    cin >> u >> v;
    --u, --v;
    edge[i] = make_pair(u, v);
    g[u].push_back(i);
  }
  for (int i = 0; i < N; ++i)
    if (dfs_num[i] != 2) dfs(i);
  cout << (cycle_found ? 2 : 1) << '\n';
  for (int i = 0; i < M; ++i) cout << col[i] + 1 << " \n"[i == M - 1];
}