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 (7.3.0) (judge verdict: Runtime Error)

icase=0
if icase==0:
    n,m=map(int,input().split())
    ab=[[0]*2 for i in range(n)]
    for i in range(n):
        ai,bi=map(int,input().split())
        ab[i][0]=m-ai
        ab[i][1]=-bi
elif icase==1:
    n,m=3,4
    ab=[[2, -2], [0, -1], [0, -3]]    
elif icase==2:
    n,m=5,3
    ab=[[1, -3], [1, -1], [2, -4], [2, -3], [2, -2]]

from heapq import heappop,heappush
#heappush(H,l) 
#print(heappop(H)) #2

#ab.sort(reverse=True)
ab.sort()
a0=ab[0][0]
h=[]
asum=0
idx=0
for i in range(m):
    while ab[idx][0]<m-i:
        heappush(h,ab[idx][1])
#        print("i:",i,"idx:",idx,h)
        idx+=1
        if idx>=n:
            asum+=heappop(h)
#            print("1:",idx,asum,h)
            break
    if idx>=n:
        asum+=heappop(h)
#        print("2:",idx,asum,h)
        break

if idx>=n:
    asum+=heappop(h)
#    print("3:",idx,asum,h)

print(-asum)