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: Accepted)

# -*- coding: utf-8 -*-
"""
Created on Sun Apr 12 02:06:07 2020

@author: Kanaru Sato
"""

import heapq

n,m = list(map(int, input().split()))
albeit = []
for i in range(m+1):
    albeit.append([])

for i in range(n):
    a,b = list(map(int, input().split()))
    if a <= m:
        albeit[a].append(-1*b)

##print(albeit)
Heap = [0]
sum = 0
for i in range(m):
    day = i+1
    for money in albeit[day]:
        heapq.heappush(Heap, money)
    if Heap[0] != 0:
        daymax = heapq.heappop(Heap)*-1
        sum += daymax

print(sum)