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: Time Limit Exceeded)

n,m=[int(i) for i in input().split()]
c={}
for _ in range(n):
  a,b=[int(i) for i in input().split()]
  if a in c:
    c[a].append(b)
  else:
    c[a]=[b]
k=0
l=[]
for i in range(m+1):
  if i in c:
    j=list(reversed(sorted(c[i])))
    if l:
      if j[0]>l[0]:
        k+=j[0]
        if len(j)>1:
          l=list(reversed(sorted(l+j[1:])))
      else:
        k+=l[0]
        l=list(reversed(sorted(l[1:]+j)))
    else:
      k+=j[0]
      l=j[1:]
  else:
    if l:
      k+=l[0]
      l=l[1:]
  if len(l)>m:
    l=l[:m]
print(k)