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)

from heapq import heappush, heappop

N, M = map(int, input().split())
jobs = [[] for _ in range(M)]

for i in range(N):
  ab = list(map(int, input().split()))
  #Ai > M の仕事はいらない
  if ab[0] <= M:
    jobs[ab[0] - 1].append(ab[1])

#最終日はAi=1の仕事のみ選べる。その前の日はAi<=2, ...
#終わりから考えればよさそう
hq = []
ans = 0
for i in range(M):
  for b in jobs[i]:
    heappush(hq, -1 * b)
  if len(hq) > 0:
    m = -1 * heappop(hq)
    ans += m

print(ans)