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 PyPy3 (2.4.0) (judge verdict: Accepted)

import sys
from heapq import heapify, heappop, heappush

sr = lambda: sys.stdin.readline().rstrip()
ir = lambda: int(sr())
lr = lambda: list(map(int, sr().split()))

N, M = lr()
jobs = [[] for _ in range(M+1)]
for _ in range(N):
    a, b = lr()
    if a > M:
        continue
    jobs[a].append(b)

answer = 0
candidate = []
for i in range(1, M+1):
    for job in jobs[i]:
        heappush(candidate, -job)
    if candidate:
        answer += heappop(candidate)

answer *= -1
print(answer)
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