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;
int n, m, VT, ans[5005], d[5005];
bool del[5005];
int dfn[5005], low[5005], scc, S[5005], tp;
bool mark[5005];
int TOT, LA[5005], NE[5005], EN[5005], ID[5005];
int tot, la[5005], ne[5005], en[5005], id[5005];
void Link(int x, int y, int i) {
  EN[++TOT] = y;
  NE[TOT] = LA[x];
  LA[x] = TOT;
  ID[TOT] = i;
  en[++tot] = x;
  ne[tot] = la[y];
  la[y] = tot;
  id[tot] = i;
}
void tarjan(int x) {
  dfn[x] = low[x] = ++VT;
  S[++tp] = x;
  mark[x] = 1;
  for (int i = LA[x]; i; i = NE[i]) {
    int y = EN[i];
    if (!dfn[y]) {
      tarjan(y);
      low[x] = min(low[x], low[y]);
    } else if (mark[y])
      low[x] = min(low[x], dfn[y]);
  }
  if (dfn[x] == low[x]) {
    int y;
    scc++;
    do {
      y = S[tp--];
      mark[y] = 0;
    } while (y != x);
  }
}
void work(int x) {
  for (int i = LA[x]; i; i = NE[i]) {
    int y = EN[i];
    if (del[y]) continue;
    ans[ID[i]] = 1;
  }
  for (int i = la[x]; i; i = ne[i]) {
    int y = en[i];
    if (del[y]) continue;
    ans[id[i]] = 2;
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    Link(x, y, i);
  }
  for (int i = 1; i <= n; i++)
    if (!dfn[i]) tarjan(i);
  if (scc == n) {
    printf("%d\n", 1);
    for (int i = 1; i <= m; i++) printf("%d ", 1);
  } else {
    printf("%d\n", 2);
    for (int i = 1; i <= n; i++) work(i);
    for (int i = 1; i <= m; i++) printf("%d ", ans[i]);
  }
}