Problem:
Score : 400 points

Problem Statement

There are N one-off jobs available. If you take the i-th job and complete it, you will earn the reward of B_i after A_i days from the day you do it.

You can take and complete at most one of these jobs in a day.

However, you cannot retake a job that you have already done.

Find the maximum total reward that you can earn no later than M days from today.

You can already start working today.

Constraints

All values in input are integers.

1 \leq N \leq 10^5

1 \leq M \leq 10^5

1 \leq A_i \leq 10^5

1 \leq B_i \leq 10^4

Input

Input is given from Standard Input in the following format:

N M
A_1 B_1
A_2 B_2
\vdots
A_N B_N

Output

Print the maximum total reward that you can earn no later than M days from today.

Sample Input 1

3 4
4 3
4 1
2 2

Sample Output 1

5

You can earn the total reward of 5 by taking the jobs as follows:

Take and complete the first job today. You will earn the reward of 3 after four days from today.

Take and complete the third job tomorrow. You will earn the reward of 2 after two days from tomorrow, that is, after three days from today.

Sample Input 2

5 3
1 2
1 3
1 4
2 1
2 3

Sample Output 2

10

Sample Input 3

1 1
2 1

Sample Output 3

0

Solution in Python (3.4.3) (judge verdict: Runtime Error)

import math
from itertools import accumulate


class Job:

    def __repr__(self):
        return f'{self.reward},{self.days}'

    def __init__(self, job_id, reward, days):
        self.id = job_id
        self.reward = reward
        self.days = days


def dfs(memo, jobs, remaing_days):
    reward = 0

    key = str(jobs)

    if key in memo[remaing_days]:
        # print('return memo')
        return memo[remaing_days][key]

    if remaing_days == 0:
        return 0

    for job in jobs:
        if job.days <= remaing_days:
            # print(
            #     f'trying {job} sum: {reward} remaing_days:{remaing_days} selected:{selected_jobs}')

            remaining_jobs = jobs - set([job])
            new_reward = job.reward + \
                dfs(memo, remaining_jobs, remaing_days - 1)

            # if reward < new_reward:
            #     print(
            #          f'pickup {job} sum: {new_reward} remaing_days:{remaing_days} selected:{selected_jobs}')
            reward = max(reward, new_reward)
        # else:
        #     print(
        #         f'fail to pickup {job} sum: {reward} remaing_days:{remaing_days} selected:{selected_jobs}')

    memo[remaing_days][key] = reward

    return reward


N, M = list(map(int, input().split()))
jobs = set()

memo = { day: {}  for day in range(M+1) }

for i in range(N):
    A, B = list(map(int, input().split()))
    jobs.add(Job(job_id=i, reward=B, days=A))

print(dfs(memo, jobs, M))