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 
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 5050;
bool circled, vis[MAXN];
int n, m, du[MAXN];
vector<int> G[MAXN];
vector<pair<int, int>> edges;
void AddEdge(int from, int to) {
  ++du[to];
  int id = edges.size();
  edges.emplace_back(from, to);
  G[from].push_back(id);
}
bool Circle() {
  int cnt = n;
  queue<int> q;
  for (int i = 0; i < n; ++i) {
    if (du[i] == 0) q.push(i);
  }
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    --cnt;
    for (int id : G[u]) {
      int v = edges[id].second;
      if (--du[v] == 0) q.push(v);
    }
  }
  return cnt > 0;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 0; i < m; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    --u, --v;
    AddEdge(u, v);
  }
  circled = Circle();
  int cnt = circled ? 2 : 1;
  printf("%d\n", cnt);
  vector<int> ans;
  if (!circled) {
    for (int i = 0; i < m; ++i) ans.push_back(1);
  } else {
    for (int i = 0; i < m; ++i) {
      auto& p = edges[i];
      if (p.first > p.second)
        ans.push_back(1);
      else
        ans.push_back(2);
    }
  }
  for (int i = 0; i < m; ++i) {
    printf("%d%c", ans[i], " \n"[i == m - 1]);
  }
}
