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)

from collections import defaultdict
N, M = map(int, input().split())
AB_dict = defaultdict(list)
AB = sorted([tuple(map(int, input().split())) for i in range(N)], reverse=True, key=lambda x: x[0])
AB = sorted(AB, reverse=True, key=lambda x: x[1])
EF = [M-i for i in range(10**5+1)]
ans = [0 for i in range(M)]
for i in range(N):
    Ai, Bi = AB[i]
    ans_i = EF[Ai]
    while ans_i >= 0 and ans[ans_i] != 0:
        ans_i -= 1
    if ans_i >= 0:
        ans[ans_i] = max(ans[ans_i], Bi)
        EF[Ai] = ans_i - 1
    else:
        EF[Ai] = ans_i
print(sum(ans))