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 nax = (long long)5e3 + 5;
const long long white = 0, gray = 1, black = 2;
bool cycle = false;
vector<pair<long long, long long>> g[nax];
vector<long long> color(nax, white), par(nax, -1), ans(nax, 1);
vector<long long> d(nax, 0), f(nax, 0);
long long n, m, timer = 0;
void dfs(long long u) {
  d[u] = timer++;
  color[u] = gray;
  for (auto p : g[u]) {
    long long v = p.first, idx = p.second;
    if (color[v] == black) {
      if (d[u] < d[v])
        ans[idx] = 1;
      else
        ans[idx] = 1;
    } else if (color[v] == gray) {
      cycle = true;
      ans[idx] = 2;
    } else if (color[v] == white) {
      par[v] = u;
      ans[idx] = 1;
      dfs(v);
    }
  }
  color[u] = black;
  f[u] = timer++;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (long long i = 0, u, v; i < m; i++) {
    cin >> u >> v;
    --u, --v;
    g[u].push_back({v, i});
  }
  for (long long i = 0; i < n; i++)
    if (color[i] == white) dfs(i);
  if (!cycle) {
    cout << 1 << '\n';
    for (long long i = 0; i < m; i++) cout << 1 << " ";
  } else {
    cout << 2 << '\n';
    for (long long i = 0; i < m; i++) cout << ans[i] << " ";
  }
  return 0;
}
