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: Accepted)

n,m=map(int,input().split())

lists=[[] for _ in range(m+1)]
#lists[i]=i日後にお金を得られる時

for _ in range(n):
    a,b=map(int,input().split())
    #a日後に報酬ｂである
    if 1<=a<=m:
        lists[a].append(b)
    else:
        pass
    
import heapq


heap=[]
heapq.heapify(heap)

ans=0

for i in range(1,m+1):
    for some in lists[i]:
        #i日後にお金を手に入れられる
        heapq.heappush(heap,-some)
    if heap:
        x=heapq.heappop(heap)
        x=x*(-1)
        ans+=x
        
print(ans)