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 long long POSITIVE_INFINITY = 9223372036854775807;
const long long NEGATIVE_INFINITY = -9223372036854775807;
const long long MOD = 1000000007;
const long double PI = acos(-1.0);
template <typename T>
inline T bigMod(T A, T B, T M = MOD) {
  A %= M;
  T ret = 1;
  while (B) {
    if (B & 1) ret = (ret * A) % M;
    A = (A * A) % M;
    B = B >> 1;
  }
  return ret;
}
template <typename T>
inline T invMod(T A, T M = MOD) {
  return bigMod(A, M - 2, M);
}
const int N = 5010;
int graph[N + 5][N + 5];
vector<int> ar[N + 5];
int color[N + 5];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int n, m;
  cin >> n >> m;
  vector<pair<int, int> > E;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    ar[x].push_back(y);
    E.push_back({x, y});
    graph[x][y] = 1;
  }
  bool f = false;
  function<void(int)> F = [&](int u) {
    color[u] = 1;
    for (int v : ar[u]) {
      if (color[v] == 1) {
        f = true;
        graph[u][v] = 2;
      }
      if (color[v] == 0) {
        F(v);
      }
    }
    color[u] = 2;
  };
  for (int i = 1; i <= n; i++) {
    if (color[i] == 0) {
      F(i);
    }
  }
  if (!f) {
    cout << 1 << "\n";
    for (int i = 0; i < m; i++) {
      cout << 1 << " ";
    }
    cout << '\n';
    ;
  } else {
    cout << 2 << '\n';
    for (int i = 0; i < m; i++) {
      int x = E[i].first;
      int y = E[i].second;
      cout << graph[x][y] << " ";
    }
    cout << '\n';
    ;
  }
  return 0;
}