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 = 5005;
vector<int> V[maxn];
int vis[maxn];
pair<int, int> edges[maxn];
bool dfs(int u) {
  vis[u] = 1;
  for (int i = 0; i < V[u].size(); i++) {
    int v = V[u][i];
    if (vis[v] == 1) {
      return true;
    } else if (vis[v] == 2) {
      continue;
    } else {
      int f = dfs(v);
      if (f) {
        return true;
      }
    }
  }
  vis[u] = 2;
  return false;
}
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  int u, v;
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &u, &v);
    V[u].push_back(v);
    edges[i] = make_pair(u, v);
  }
  bool flag = false;
  for (int i = 1; i <= n; i++) {
    if (vis[i] == 0) {
      flag |= dfs(i);
      if (flag) {
        break;
      }
    }
  }
  if (!flag) {
    puts("1");
    for (int i = 1; i <= m; i++) {
      printf("1 ");
    }
    puts("");
  } else {
    puts("2");
    for (int i = 1; i <= m; i++) {
      int val = 1;
      if (edges[i].first > edges[i].second) {
        val = 2;
      }
      printf("%d ", val);
    }
    puts("");
  }
  return 0;
}