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 N = 5007, mod = 998244353;
int n, m;
pair<int, int> a[N];
vector<int> v[N];
bool f[N];
int d[N];
int main() {
  cin >> n >> m;
  int flag = 0;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a[i].first, &a[i].second);
    v[a[i].first].push_back(a[i].second);
    d[a[i].second]++;
  }
  for (int i = 1; i <= n; i++) {
    int mn = 0;
    for (int j = 1; j <= n; j++) {
      if (d[j] == 0 && f[j] == 0) mn = j;
    }
    f[mn] = 1;
    if (mn == 0) {
      flag = 1;
      break;
    }
    for (int j = 0; j < v[mn].size(); j++) {
      d[v[mn][j]]--;
    }
  }
  if (flag == 0) {
    cout << 1 << '\n';
    for (int i = 1; i <= m; i++) {
      printf("1 ");
    }
  } else {
    cout << 2 << '\n';
    for (int i = 1; i <= m; i++) {
      if (a[i].first < a[i].second)
        printf("1 ");
      else
        printf("2 ");
    }
  }
  return 0;
}
