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;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    if (ch == '-') f = -1;
  for (; ch >= '0' && ch <= '9'; ch = getchar())
    x = (x << 1) + (x << 3) + ch - '0';
  return x * f;
}
const int N = 5005;
vector<pair<int, int> > G[N];
int ans[N], vis[N], in[N], u[N], v[N], num = 1;
void dfs(int u) {
  vis[u] = 1;
  in[u] = 1;
  for (auto i : G[u])
    if (in[i.first])
      ans[i.second] = 2, num = 2;
    else {
      ans[i.second] = 1;
      if (!vis[i.first]) dfs(i.first);
    }
  in[u] = 0;
}
int main() {
  int n = read();
  int m = read();
  for (int i = 1; i <= m; ++i) {
    u[i] = read();
    v[i] = read();
    G[u[i]].push_back({v[i], i});
  }
  for (int i = 1; i <= n; ++i)
    if (!vis[i]) dfs(i);
  printf("%d\n", num);
  for (int i = 1; i <= m; ++i) printf("%d ", ans[i]);
  return 0;
}
