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())
A = [0] * N
B = [0] * N
ab = []
for i in range(N):
    A[i], B[i] = map(int, input().split())
    ab.append([A[i], B[i]])

ab.sort()
i = N
ans = 0
nums = []
while 1 <= i:
    if M <= 0:
        nums.sort()
        nums.reverse()
        if nums[-1] < ab[i-1][1]:
            nums.pop()
            nums.append(ab[i-1][1])
    if M >= 1:
        if ab[i-1][0] <= M:
            nums.append(ab[i-1][1])
        else:
            while ab[i-1][0] > M:
                i -= 1
                if i - 1 < 0:
                    break
                nums.append(ab[i-1][1])
    i -= 1
    M -= 1

print(sum(nums))