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 int mod = 1e9 + 7;
double rast(int x1, int y1, int x2, int y2) {
  return sqrt((x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1));
}
vector<int> g[10000];
bool has_cycle = false;
int used[10005];
void dfs(int v) {
  used[v] = 1;
  for (int to : g[v]) {
    if (used[to] == 1)
      has_cycle = true;
    else if (used[to] == 0)
      dfs(to);
  }
  used[v] = 2;
}
int main() {
  ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > edges;
  for (int i = 0; i < m; i++) {
    int l, r;
    cin >> l >> r;
    l--, r--;
    edges.push_back({l, r});
    g[l].push_back(r);
  }
  for (int i = 0; i < n; i++)
    if (!used[i]) dfs(i);
  if (!has_cycle) {
    cout << 1 << '\n';
    for (int i = 0; i < m; i++) cout << 1 << " ";
  } else {
    cout << 2 << '\n';
    for (int i = 0; i < m; i++)
      if (edges[i].first < edges[i].second)
        cout << 1 << " ";
      else
        cout << 2 << " ";
  }
}
