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 MOD = 1000000007;
int fx[] = {0, 0, -1, 1, +1, -1, -1, +1};
int fy[] = {1, -1, 0, 0, +1, +1, -1, -1};
bool sortinrev(const pair<int, int> &a, const pair<int, int> &b) {
  return a.first > b.first;
}
template <typename T>
inline T Bigmod(T base, T power, T MOD) {
  T ret = 1;
  while (power) {
    if (power & 1) ret = (ret * base) % MOD;
    base = (base * base) % MOD;
    power >>= 1;
  }
  return ret;
}
const int N = 5005;
int color[N];
int res[N];
bool cyc;
vector<pair<int, int>> adj[N];
void init() {
  memset(color, 0, sizeof(color));
  memset(res, 0, sizeof(res));
  cyc = 0;
  for (int i = 0; i < N; i++) adj[i].clear();
}
void dfs(int src) {
  color[src] = 1;
  for (int v = 0; v < adj[src].size(); v++) {
    int to = adj[src][v].first, id = adj[src][v].second;
    if (color[to] == 0) {
      dfs(to);
      res[id] = 1;
    } else if (color[to] == 2) {
      res[id] = 1;
    } else {
      cyc = 1;
      res[id] = 2;
    }
  }
  color[src] = 2;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ;
  int n, m;
  while (cin >> n >> m) {
    init();
    for (int i = 0; i < m; i++) {
      int u, v;
      cin >> u >> v;
      u--, v--;
      adj[u].push_back(make_pair(v, i));
    }
    for (int i = 0; i < n; i++) {
      if (color[i] == 0) {
        dfs(i);
      }
    }
    if (cyc) {
      cout << 2 << endl;
      for (int i = 0; i < m; i++) {
        cout << res[i] << " ";
      }
      cout << endl;
    } else {
      cout << 1 << endl;
      for (int i = 0; i < m; i++) cout << 1 << " ";
      cout << endl;
    }
  }
  return 0;
}