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 MAX = 5000;
int head[MAX + 10], nxt[MAX + 10], to[MAX + 10], clr[MAX + 10], tot;
bool vis[MAX + 10], is[MAX + 10];
void dfs(int x) {
  vis[x] = is[x] = true;
  for (int e = head[x]; e; e = nxt[e]) {
    if (!vis[to[e]])
      clr[e] = 1;
    else
      clr[e] = 2;
    if (!is[to[e]]) dfs(to[e]);
  }
  vis[x] = false;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n, m, u, v;
  cin >> n >> m;
  for (int i = 1; i <= m; ++i) {
    cin >> u >> v;
    nxt[++tot] = head[u];
    to[head[u] = tot] = v;
  }
  for (int i = 1; i <= n; ++i) {
    if (!is[i]) dfs(i);
  }
  int k = 0;
  for (int i = 1; i <= m; ++i) {
    k = max(k, clr[i]);
  }
  cout << k << '\n';
  for (int i = 1; i <= m; ++i) {
    cout << clr[i] << ' ';
  }
  return 0;
}
