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 Python (3.4.3) (judge verdict: Runtime Error)

import numpy as np

temp=list(map(int, input().split()))
N=temp[0]
M=temp[1]

AB=np.array([list(map(int, input().split())) for i in range(N)])
AB=np.c_[AB, np.zeros((AB.shape[0], 1))]

money=0
for i in range(1, M+1):
  temp=AB[AB[:, 0]<=i]
  temp=temp[temp[:, 2]==0]
  temp=temp[np.argmax(temp[:, 1])]
  money+=temp[1]
  
  x=0
  while x<AB.shape[0]:
    if sum(AB[x, :]==temp)==3:
      AB[x, 2]=1
      break
    x+=1
    
print(int(money))