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 sys
import heapq
imput = sys.stdin.readline

def main():
    n,m = map(int,input().split())
    money = [[] for _ in range(10**5)]
    dp = [0] * m
    hq = []
    heapq.heapify(hq)

    for _ in range(n):
        a,b = map(int,input().split())
        money[a-1] += [b]

    for i in range(m):
        for j in money[i]:
            heapq.heappush(hq,-j)
        if hq:
            dp[i] += -heapq.heappop(hq)
        if i >= 1:
            dp[i] += dp[i-1]
    print(dp[-1])

if __name__ == '__main__':
    main()