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;
vector<vector<int>> ng;
map<pair<int, int>, int> mp;
int vis[5005];
int flag = 0;
void dfs(int u, int p) {
  pair<int, int> temp = make_pair(p, u);
  if (vis[u] == 1) {
    mp[temp] = 2;
    flag = 1;
    return;
  }
  if (vis[u] == 2) {
    mp[temp] = 1;
    return;
  }
  if (p != -1) mp[temp] = 1;
  vis[u] = 1;
  for (auto x : ng[u]) {
    dfs(x, u);
  }
  vis[u] = 2;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  int m;
  cin >> n >> m;
  ng.resize(n);
  vector<pair<int, int>> edg(m);
  int u, v;
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    edg[i] = make_pair(u - 1, v - 1);
    ng[u - 1].push_back(v - 1);
  }
  for (int i = 0; i < n; i++) {
    if (vis[i] == 0) dfs(i, -1);
  }
  if (flag == 0)
    cout << 1 << endl;
  else
    cout << 2 << endl;
  for (int i = 0; i < m; i++) {
    cout << mp[edg[i]] << " ";
  }
  cout << endl;
  return 0;
}
