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 = 5005;
vector<int> G[maxn];
int id[maxn][maxn];
void insert(int u, int v, int i) {
  G[u].push_back(v);
  id[u][v] = i;
}
int vis[maxn], col[maxn];
int cyc = 0;
void dfs(int u) {
  vis[u] = 1;
  for (int v : G[u]) {
    if (!vis[v]) {
      col[id[u][v]] = 1;
      dfs(v);
    } else if (vis[v] == 1) {
      col[id[u][v]] = 2;
      cyc = 1;
    } else {
      col[id[u][v]] = 1;
    }
  }
  vis[u] = 2;
}
int n, m;
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1, u, v; i <= m; i++) {
    scanf("%d%d", &u, &v);
    insert(u, v, i);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i);
  if (cyc)
    puts("2");
  else
    puts("1");
  for (int i = 1; i <= m; i++) printf("%d ", col[i]);
  return 0;
}
