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)

n, m = map(int, input().split())
AB = []
for i in range(n):
    a, b  = map(int, input().split())
    if b <= m:
        AB.append((a, b))
AB.sort(key=lambda ab: -ab[1])
counts = [0] * m
cands = [0] * m
total = 0
for a, b in AB:
    if a > m:
        continue
    for i in range(a - 1 + counts[a - 1], m):
        if cands[i] > 0:
            continue
        cands[i] = b
        counts[a - 1] += 1
        total += 1
        break
    if total == m:
        break
print(sum(cands))