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 deg[5001];
vector<int> redge[5001];
int main() {
  int n, m;
  cin >> n >> m;
  vector<int> ans;
  for (int i = 1; i <= m; ++i) {
    int x, y;
    cin >> x >> y;
    ans.push_back((x < y) + 1);
    ++deg[x];
    redge[y].push_back(x);
  }
  vector<int> st;
  for (int i = 1; i <= n; ++i) {
    if (deg[i] == 0) st.push_back(i);
  }
  for (int i = 0; i < st.size(); ++i) {
    int x = st[i];
    for (int j : redge[x]) {
      if (--deg[j] == 0) st.push_back(j);
    }
  }
  if (st.size() < n) {
    printf("2\n");
    for (int i : ans) printf("%d ", i);
  } else {
    printf("1\n");
    for (int i = 0; i < m; ++i) printf("1 ");
  }
  return 0;
}
