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;
bool vis[5001];
int x1 = 0;
bool vis2[5001];
vector<int> g[5001];
bool dfs(int s) {
  vis[s] = true;
  x1++;
  vis2[s] = true;
  for (auto it : g[s]) {
    if (!vis[it]) return dfs(it);
    if (vis2[it]) return true;
  }
  return false;
}
void solve() {
  int i, j = 0, r = 200, p, q, k, d, y = 0, h, flag = 0, n, m = 0,
         a = 1000000007, b = 1000000007, id, c, h1, h2, a1, b1, a2, w, x;
  cin >> n >> m;
  memset(vis, false, sizeof(vis));
  vector<pair<int, int> > vp;
  for (i = 0; i < m; i++) {
    cin >> a >> b;
    g[a].push_back(b);
    vp.push_back({a, b});
  }
  for (i = 1; i < n + 1; i++) {
    if (!vis[i]) {
      x1 = 0;
      memset(vis2, false, sizeof(vis2));
      if (dfs(i)) {
        flag = 1;
        break;
      }
    }
  }
  if (!flag) {
    cout << 1 << endl;
    for (i = 0; i < m; i++) cout << 1 << " ";
    return;
  }
  cout << 2 << endl;
  for (i = 0; i < m; i++) {
    if (vp[i].first < vp[i].second)
      cout << 1 << "  ";
    else
      cout << 2 << " ";
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) solve();
}
