Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5000;
const int MAXM = 5000;
pair<int, int> edge[1 + MAXM];
vector<int> g[1 + MAXN];
int inDegree[1 + MAXN], Queue[1 + MAXN];
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    g[a].push_back(b);
    edge[i] = make_pair(a, b);
    inDegree[b]++;
  }
  int start = 1, finish = 0;
  for (int i = 1; i <= n; i++)
    if (inDegree[i] == 0) {
      finish++;
      Queue[finish] = i;
    }
  while (start <= finish) {
    int node = Queue[start];
    start++;
    for (auto &it : g[node]) {
      inDegree[it]--;
      if (inDegree[it] == 0) {
        finish++;
        Queue[finish] = it;
      }
    }
  }
  if (finish == n) {
    printf("1\n");
    for (int i = 1; i <= m; i++) printf("1 ");
  } else {
    printf("2\n");
    for (int i = 1; i <= m; i++)
      if (edge[i].first < edge[i].second)
        printf("1 ");
      else
        printf("2 ");
  }
  return 0;
}