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: Wrong Answer)

import heapq

N, M = [int(_) for _ in input().split(' ')]
ABs = [[] for _ in range(M + 1)]
r = 0

for i in range(N):
    A, B = [int(x) for x in input().split()]
    if A <= M:
        ABs[A].append(-B)

for i in range(M + 1):
    heapq.heapify(ABs[i])
    print("%d: %s" % (i, str(ABs[i])))

for m in range(M, 0, -1):
    B_max = -1
    A_max = -1

    for a in range(m, 0, -1):
        if len(ABs[a]) == 0:
            continue

        b = - ABs[a][0]
        if B_max < b:
            A_max = a
            B_max = b

    if A_max != -1:
        r -= heapq.heappop(ABs[A_max])

print(r)