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;
int n, m, u, v, num_col = 1;
vector<int> adjList[5001];
vector<int> edges[5001];
bitset<5001> vis;
bitset<5001> act;
bitset<5001> col;
void dfs(int nd) {
  vis[nd] = 1;
  act[nd] = 1;
  for (int k = 0; k < adjList[nd].size(); k++) {
    int nd_b = adjList[nd][k];
    int edge_num = edges[nd][k];
    if (!vis[nd_b])
      dfs(nd_b);
    else if (act[nd_b]) {
      num_col = 2;
      col[edge_num] = 1;
    }
  }
  act[nd] = 0;
}
int main() {
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> u >> v;
    u--;
    v--;
    adjList[u].push_back(v);
    edges[u].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    if (!vis[i]) dfs(i);
  }
  cout << num_col << endl;
  for (int i = 0; i < m; i++) {
    cout << (((int)col[i]) + 1) << (i < m - 1 ? " " : "");
  }
  cout << endl;
  return 0;
}
