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: Runtime Error)

N,M = map(int,input().split())

A = [0]*N
B = [0]*N

for n in range(N):
  B[n],A[n] = map(int,input().split())
  
CA = CA
CB = CB
CN = N

#F = [True]*N
sr = 0
for m in range(M):
  cr = 0
#  ci = -1
  CN = 0
  for n in range(CN):
    if CB[n] <= m+1  and cr<CA[n]:
      #ci = n
      cr = A[n]
    else:
      CA[CN] = A[n]
      CB[CN] = B[n]
      CN += 1
    sr += cr
#  if ci!=-1:
#    sr += cr
#    F[ci] = False
print(sr)