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 N = 5005;
int n, m, u[N], v[N];
vector<int> adj[N];
vector<int> topsort;
int pos[N];
bool visited[N];
void dfs(int u) {
  if (visited[u]) return;
  visited[u] = true;
  for (int v : adj[u]) {
    dfs(v);
  }
  topsort.push_back(u);
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  for (int i = 0; i < m; i++) {
    cin >> u[i] >> v[i];
    adj[u[i]].push_back(v[i]);
  }
  for (int i = 1; i <= n; i++) {
    dfs(i);
  }
  for (int i = 0; i < n; i++) {
    pos[topsort[i]] = i;
  }
  bool cycle = false;
  for (int i = 0; i < m; i++) {
    if (pos[u[i]] < pos[v[i]]) {
      cycle = true;
    }
  }
  if (!cycle) {
    cout << 1 << endl;
    for (int i = 0; i < m; i++) {
      cout << 1 << " ";
    }
    cout << endl;
  } else {
    cout << 2 << endl;
    for (int i = 0; i < m; i++) {
      cout << (u[i] < v[i] ? 1 : 2) << " ";
    }
    cout << endl;
  }
}
