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: Time Limit Exceeded)

from functools import cmp_to_key
def cmp(x, y):
  if x[0] != y[0]:
    return y[0] - x[0]
  else:
    return y[1] - x[1]
n, m = map(int, input().split())
data = [[int(e) for e in input().split()] for _ in range(n)]
data.sort(key = cmp_to_key(cmp))
t = [0] * (m + 1)
for i in range(n):
  a, b = data[i]
  for j in range(m, -1, -1):
    if j + a <= m:
      t[j + 1] = max(t[j + 1], t[j] + b)
print(max(t))