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;
struct edge {
  int to;
  int num;
};
const int maxn = 5123;
vector<edge> G[maxn];
int color[maxn];
int visit[maxn];
int v = -1;
int flag = 0;
void dfs(int node) {
  visit[node] = 1;
  for (int i = 0; i < G[node].size(); i++) {
    edge e = G[node][i];
    if (color[e.num] == 0) {
      if (visit[e.to] == 0) {
        color[e.num] = 1;
        dfs(e.to);
      } else {
        flag = 1;
        color[e.num] = 2;
      }
    }
  }
  visit[node] = 0;
}
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  for (int i = 1; i <= m; i++) {
    int a, b;
    cin >> a >> b;
    edge e;
    e.to = b, e.num = i;
    G[a].push_back(e);
  }
  memset(color, 0, sizeof(color));
  memset(visit, 0, sizeof(visit));
  for (int i = 1; i <= n; i++) {
    if (visit[i] == 0) {
      dfs(i);
    }
  }
  if (flag == 0)
    cout << "1" << endl;
  else
    cout << "2" << endl;
  for (int i = 1; i <= m; i++) {
    cout << color[i] << " ";
  }
  cout << endl;
  return 0;
}
