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)

from sys import stdin
def main()
  from builtins import int, map, range
  readline = stdin.readline
  from heapq import heappush, heappop
  n, m = map(int, readline().split())
  jobs = {}
  for _ in range(n):
    a, b = map(int, readline().split())
    if a > m:
      continue
    if a in jobs:
      jobs[a].append(b)
    else:
      jobs[a] = [b]
  result = 0
  candidates = []
  for a in range(1, m + 1):
    if a in jobs:
      for b in jobs[a]:
        heappush(candidates, -b)
    else:
      if len(candidates) == 0:
        continue
    result += -heappop(candidates)
  print(result)
main()