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;
const long long MOD = 1e9 + 7;
class {
 public:
  int n, m;
  vector<pair<int, int> > V[5555];
  bool vis[5555], in[5555];
  int ans[5555];
  void solve() {
    cin >> n >> m;
    for (int i = 0; i < m; ++i) {
      int u, v;
      cin >> u >> v;
      V[u].push_back({v, i});
    }
    for (int i = 1; i <= n; ++i) {
      if (!vis[i]) dfs(i);
    }
    if (two)
      cout << 2 << "\n";
    else
      cout << 1 << "\n";
    for (int i = 0; i < m; ++i) {
      cout << ans[i] + 1 << " ";
    }
  }
  bool two = 0;
  void dfs(int u) {
    in[u] = 1;
    vis[u] = 1;
    for (auto v : V[u]) {
      if (!vis[v.first]) {
        dfs(v.first);
      } else if (in[v.first]) {
        ans[v.second] = 1;
        two = 1;
      }
    }
    in[u] = 0;
  }
} NSPACE;
int main() {
  ios_base::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  NSPACE.solve();
}
