InternBootcamp/internbootcamp/bootcamp/ccoffeebreak/ccoffeebreak.py

"""# 

### 谜题描述
Recently Monocarp got a job. His working day lasts exactly m minutes. During work, Monocarp wants to drink coffee at certain moments: there are n minutes a_1, a_2, ..., a_n, when he is able and willing to take a coffee break (for the sake of simplicity let's consider that each coffee break lasts exactly one minute). 

However, Monocarp's boss doesn't like when Monocarp takes his coffee breaks too often. So for the given coffee break that is going to be on minute a_i, Monocarp must choose the day in which he will drink coffee during the said minute, so that every day at least d minutes pass between any two coffee breaks. Monocarp also wants to take these n coffee breaks in a minimum possible number of working days (he doesn't count days when he is not at work, and he doesn't take coffee breaks on such days). Take into account that more than d minutes pass between the end of any working day and the start of the following working day.

For each of the n given minutes determine the day, during which Monocarp should take a coffee break in this minute. You have to minimize the number of days spent. 

Input

The first line contains three integers n, m, d (1 ≤ n ≤ 2⋅10^{5}, n ≤ m ≤ 10^{9}, 1 ≤ d ≤ m) — the number of coffee breaks Monocarp wants to have, the length of each working day, and the minimum number of minutes between any two consecutive coffee breaks.

The second line contains n distinct integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ m), where a_i is some minute when Monocarp wants to have a coffee break.

Output

In the first line, write the minimum number of days required to make a coffee break in each of the n given minutes. 

In the second line, print n space separated integers. The i-th of integers should be the index of the day during which Monocarp should have a coffee break at minute a_i. Days are numbered from 1. If there are multiple optimal solutions, you may print any of them.

Examples

Input

4 5 3
3 5 1 2


Output

3
3 1 1 2 


Input

10 10 1
10 5 7 4 6 3 2 1 9 8


Output

2
2 1 1 2 2 1 2 1 1 2 

Note

In the first example, Monocarp can take two coffee breaks during the first day (during minutes 1 and 5, 3 minutes will pass between these breaks). One break during the second day (at minute 2), and one break during the third day (at minute 3).

In the second example, Monocarp can determine the day of the break as follows: if the minute when he wants to take a break is odd, then this break is on the first day, if it is even, then this break is on the second day.

Here is a reference code to solve this task. You can use this to help you genereate cases or validate the solution.
```python
from __future__ import print_function
from heapq import *


def read():
    return [int(num) for num in raw_input().split()]


n, m, d = read()
tm_raw = read()

tm = list(sorted(tm_raw))

tail_tm = []
day_id = {}

for t in tm:
    if len(tail_tm) > 0 and t - tail_tm[0][0] > d:
        day_id[t] = tail_tm[0][1]
        heapreplace(tail_tm, (t, tail_tm[0][1]))
    else:
        day_id[t] = len(tail_tm) + 1
        heappush(tail_tm, (t, len(tail_tm) + 1))

print(len(tail_tm))
print(' '.join([str(day_id[t]) for t in tm_raw]))
```


请完成上述谜题的训练场环境类实现，包括所有必要的方法。
"""

from bootcamp import Basebootcamp
import random
from heapq import heappush, heappop
import re
from collections import defaultdict
from bootcamp import Basebootcamp

class Ccoffeebreakbootcamp(Basebootcamp):
    def __init__(self, max_n=10, max_m=100):
        self.max_n = max_n
        self.max_m = max_m

    def case_generator(self):
        n = random.randint(1, self.max_n)
        m = random.randint(n, self.max_m)
        a = random.sample(range(1, m + 1), n)
        d = random.randint(1, m)
        return {
            'n': n,
            'm': m,
            'd': d,
            'a': a
        }

    @staticmethod
    def prompt_func(question_case):
        n = question_case['n']
        m = question_case['m']
        d = question_case['d']
        a_list = question_case['a']
        a_str = ' '.join(map(str, a_list))
        prompt = f"""Monocarp wants to schedule his coffee breaks during his working day, which lasts for {m} minutes. He has specified {n} distinct minutes during which he would like to have coffee. However, his boss requires that any two consecutive coffee breaks on the same day must be at least {d} minutes apart. Your task is to determine the minimum number of days required and assign each coffee break to a specific day, ensuring that the breaks on the same day are spaced appropriately. Days are numbered starting from 1.

The coffee break minutes are: {a_str}

Your answer must be formatted as follows:

[answer]
<minimum number of days>
<space-separated day assignments in the order of the input>
[/answer]

For example, if the minimum days is 3 and the assignments are 1, 2, 1, the answer should be:

[answer]
3
1 2 1
[/answer]

Please ensure the answer is enclosed within [answer] and [/answer] tags."""
        return prompt

    @staticmethod
    def extract_output(output):
        matches = re.findall(r'\[answer\](.*?)\[/answer\]', output, re.DOTALL)
        if not matches:
            return None
        answer_block = matches[-1].strip()
        lines = [line.strip() for line in answer_block.split('\n') if line.strip()]
        if len(lines) != 2:
            return None
        try:
            k = int(lines[0])
            days = list(map(int, lines[1].split()))
        except:
            return None
        if len(days) == 0:
            return None
        return (k, days)

    @classmethod
    def _verify_correction(cls, solution, identity):
        if solution is None:
            return False
        user_k, user_days = solution
        a = identity['a']
        n = identity['n']
        d = identity['d']
        m = identity['m']

        tm = sorted(a)
        tail_tm = []
        day_id = {}
        for t in tm:
            if tail_tm and t - tail_tm[0][0] > d:
                existing_day = tail_tm[0][1]
                day_id[t] = existing_day
                heappop(tail_tm)
                heappush(tail_tm, (t, existing_day))
            else:
                new_day = len(tail_tm) + 1
                day_id[t] = new_day
                heappush(tail_tm, (t, new_day))
        k_correct = len(tail_tm)
        days_correct = [day_id[t] for t in a]

        if user_k != k_correct:
            return False

        if len(user_days) != n:
            return False

        if any(day < 1 or day > user_k for day in user_days):
            return False

        day_groups = defaultdict(list)
        for time, day in zip(a, user_days):
            day_groups[day].append(time)

        for times in day_groups.values():
            sorted_times = sorted(times)
            for i in range(1, len(sorted_times)):
                if sorted_times[i] - sorted_times[i-1] < d:
                    return False
        return True