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 heapq

n,m = map(int, input().split())
ab = [list(map(int, input().split())) for _ in range(n)]
ab.append([m+1,0])
ab = sorted(ab, key=lambda x:x[0])



max_values = [0 for _ in range(m)]
hq = []

possible_days = 0

for now in range(n+1):
    # if necessary day exceed the (now_day), extract the most valuable work
    while possible_days <= ab[now][0]:
        if len(hq) < 1:
            pass
        else:
            max_values[possible_days-2] = -heapq.heappop(hq)
        possible_days += 1
    if possible_days-2 > m-1:
        break
    heapq.heappush(hq, -ab[now][1])
print(sum(max_values))