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>
constexpr auto MAXN = 5000 << 1;
constexpr auto MAXM = MAXN;
short inque[MAXN], color[MAXM];
bool cyc = false;
struct Edge {
  int to, ID;
  Edge() {}
  Edge(const int& ito, const int& iID) : to(ito), ID(iID) {}
};
std::vector<Edge> G[MAXN];
void Dfs(const int nowAt) {
  inque[nowAt] = true;
  for (const auto& it : G[nowAt])
    if (inque[it.to] == -1) {
      Dfs(it.to);
      color[it.ID] = 1;
    } else if (inque[it.to] == false) {
      color[it.ID] = 1;
    } else {
      cyc = true;
      color[it.ID] = 2;
    }
  inque[nowAt] = false;
}
signed main(void) {
  memset(inque, -1, sizeof(inque));
  int totNodes, totEdges;
  int fr, to;
  scanf("%d%d", &totNodes, &totEdges);
  for (int i = 0; i < totEdges; i++) {
    scanf("%d%d", &fr, &to);
    --fr;
    --to;
    G[fr].push_back(Edge(to, i));
  }
  for (int i = 0; i < totNodes; i++)
    if (inque[i] == -1) Dfs(i);
  puts(cyc ? "2" : "1");
  for (int i = 0; i < totEdges; i++) printf("%d ", color[i]);
  putchar('\n');
  return 0;
}
