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

w=sorted(w)
#print(w)

def solve():
  total=0
  for i in range(m,-1,-1):
#  print("i:"+str(i))
    max=0
    max_j=0
    found=False
    for j in range(len(w)):
      if (m-i)<w[j][0]: #残り日数(m-i)よりも大きい
#      print("残り日数:"+str(m-i) + "　仕事日数:"+str(w[j][0]))
        break
      elif max < w[j][1]:
#        print("最大値:"+str(w[j][1]) + " 添え字:"+str(j))
        max=w[j][1]
        max_j=j
        found=True
#  print(str(max) + "の仕事をやる")
    total = total + max
    if found:
#    print(str(max_j)+"番目の仕事を削除")
      w.pop(max_j)
#    print(w)
  return total
      
    
print(solve())