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: Time Limit Exceeded)

from heapq import heappop, heappush
n, m = map(int, input().split())
lst = [list(map(int, input().split())) for i in range(n)]
lst_f = [[] for i in range(m)]
#lst = sorted(lst, key=lambda x:x[0])
#lst = sorted(lst, key=lambda x:x[1], reverse=True)
cnt = 0
ans = 0
cntup = m
for a, b in lst:
  if a <= m:
    lst_f[a-1].append(b)
  #print(filter(lambda y:y<=m-i, map(lambda x:x[0], lst)))
  #lst2 = list(filter(lambda y:y[1]<=m-i, lst))
avb = [] 
for i in lst_f:
  for j in i:
    avb.append(-j)
  if avb != []:
    ans -= avb.pop(avb.index(min(avb)))
print(ans)