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)

N, M = map(int,input().split())
A_B = [list(map(int, input().split())) for i in range(N)]

#日数ごとの集合を作る
jobs = {m: [] for m in range(1, M + 1)}
for ab in A_B:
  if ab[0] in jobs:
    jobs[ab[0]].append(ab[1])
#print(jobs)

#日数の少ない仕事からループして最適解を見つける
import heapq
q = []
ans = 0
#仕事にかかる日数のループであって経過日のループではない
for m in range(1, M + 1):
    for b in jobs[m]:
      #qには小さい順で格納される
      heapq.heappush(q, -b)
    #print(q)
    #qが空でないとき
    if len(q) > 0:
      #現時点での日数以下の仕事で一番稼げるものを取り出す
      neg_b = heapq.heappop(q)
      ans += -neg_b

print(ans)