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;
int n, m;
int a[5001][5001], visited[5001] = {0};
long long cyc = 1;
int k = 1;
int dfs2(int node, int path[]) {
  visited[node] = 1;
  path[node] = 1;
  for (int i = 0; i < 5001; i++) {
    if (a[node][i] != 0 && path[i] == 1) {
      a[node][i] = 2;
      k = 2;
    } else if (a[node][i] != 0 && visited[i] == 0) {
      dfs2(i, path);
    }
  }
  path[node] = 0;
  return 0;
}
int dfs(int node, int curr) {
  visited[node] = cyc;
  for (int i = 0; i < n; i++) {
    if (a[node][i] != 0 && visited[i] != 0 && visited[i] == visited[node]) {
      a[node][i] = 2;
      visited[i] = -1;
      visited[node] = -1;
      k = 2;
    } else if (a[node][i] != 0 && visited[i] == -1) {
      a[node][i] = 2;
      visited[node] = -1;
      k = 2;
    } else if (a[node][i] != 0 && visited[i] != 0 &&
               visited[i] != visited[node]) {
      a[node][i] = curr;
    } else if (a[node][i] != 0 && visited[i] == 0) {
      if (visited[node] != -1) visited[node] = cyc;
      int temp = dfs(i, curr);
      if (temp == -1) visited[node] = -1;
      if (visited[node] != -1) visited[node] = cyc;
    }
  }
  if (visited[node] != -1) visited[node] = -2;
  return visited[node];
}
int main() {
  cin >> n >> m;
  memset(a, 0, sizeof(a));
  int ans = 1;
  vector<pair<int, int>> edge;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    edge.push_back(make_pair(x - 1, y - 1));
    a[x - 1][y - 1] = 1;
  }
  int path[5001] = {0};
  for (int i = 0; i < 5001; i++) {
    if (visited[i] == 0) dfs2(i, path);
  }
  cout << k << endl;
  for (int i = 0; i < m; i++) cout << a[edge[i].first][edge[i].second] << " ";
  cout << endl;
  return 0;
}