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 = []
for i in range(N):
    a, b = map(int, input().split())
    A.append([a, b])
A = sorted(A, key = lambda x : x[0], reverse = True)
# print(A)
ans = 0

for i in range(M - 1, -1, -1):
    maximum_rwd = 0
    maximum_id = 0
    for j in range(len(A)):
        if A[j][0] > M - i:
            break
        # A[j]報酬までの時間、
        elif maximum_rwd < A[j][1]:
            maximum_id = j
            maximum_rwd = A[j][1]
    if maximum_rwd > 0:
        ans += maximum_rwd
        A.pop(maximum_id)
print(ans)