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 n, m;
const int N = 1e5 + 7;
vector<vector<pair<int, int> > > g(N);
int col[N], far[N];
bool cyc;
void dfs(int u) {
  far[u] = 1;
  for (auto &i : g[u]) {
    int v = i.first, id = i.second;
    if (far[v] == 0) {
      dfs(v);
      col[id] = 1;
    } else if (far[v] == 2) {
      col[id] = 1;
    } else {
      col[id] = 2;
      cyc = true;
    }
  }
  far[u] = 2;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 0; i < m; ++i) {
    int u, v;
    cin >> u >> v;
    g[u - 1].push_back({v - 1, i});
  }
  for (int i = 0; i < n; ++i) {
    if (!far[i]) dfs(i);
  }
  cout << 1 + cyc << endl;
  for (int i = 0; i < m; ++i) cout << col[i] << ' ';
  return 0;
}
