Problem: 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 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 5e3 + 10;
int cor[maxn], pai[maxn], edge_pai[maxn], firstCycle, lastCycle;
int mark[maxn], aresta;
bool edge_mark[maxn];
vector<pair<int, int> > grafo[maxn];
bool dfs(int u) {
  mark[u] = 1;
  for (auto pp : grafo[u]) {
    int v = pp.first, e = pp.second;
    if (mark[v] == 0) {
      pai[v] = u;
      edge_pai[v] = e;
      if (dfs(v)) return true;
    } else if (mark[v] == 1 && !edge_mark[e]) {
      aresta = e;
      firstCycle = v;
      lastCycle = u;
      return 1;
    }
  }
  mark[u] = 2;
  return 0;
}
int main(void) {
  int n, m;
  scanf("%d %d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    grafo[u].push_back({v, i});
  }
  bool ok = 1;
  while (true) {
    memset(mark, 0, sizeof mark);
    bool teve = 0;
    for (int i = 1; i <= n; i++) {
      aresta = -1;
      dfs(i);
      if (aresta != -1) {
        ok = 0;
        edge_mark[aresta] = 1;
        cor[aresta] = 2;
        teve = 1;
        for (int v = lastCycle; v != firstCycle; v = pai[v])
          edge_mark[edge_pai[v]] = 1;
        break;
      }
    }
    if (!teve) break;
  }
  if (ok) {
    printf("1\n");
    for (int i = 1; i <= m; i++) printf("1 ");
    printf("\n");
    return 0;
  }
  printf("2\n");
  for (int i = 1; i <= m; i++) {
    if (cor[i] == 2)
      printf("2 ");
    else
      printf("1 ");
  }
  printf("\n");
}