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)

N,M = map(int,input().split())
#k,x = map(int,input().split())

AB=[]
for i in range(N):
    AB.append(list(map(int,input().split())))

BA=sorted(AB, key=lambda x: (x[1],x[0]))

scadule = [0]*M
full=0
for work in reversed(BA):
    if work[0]>M:
        continue
    try:
        day = scadule[work[0]-1:].index(0)
        scadule[day+work[0]-1] = work[1]
        full+=1
#        print(tmp_li,work,day,scadule)
    except:
        continue
    if full > M:
        break

ans=sum(scadule)

print(str(ans))