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 (3.8.2) (judge verdict: Wrong Answer)

from collections import defaultdict as dd
from collections import deque

n,m=map(int,input().split())
dic=dd(list)
for i in range(n):
  a,b=map(int,input().split())
  dic[a].append(b)

for k, v in dic.items():
  dic[k]=sorted(v, reverse=True)
  dic[k]=deque(dic[k])

ans=0
flg=[False]*(m+1)
for i in range(m+1):
  if i in dic.keys():
    flg[m-i]=True
    ans+=dic[i][0]
    dic[i].popleft()

    
for i in range(m):
  if not flg[i]:
    tv=tk=0
    for k, v in dic.items():
      if len(v)==0: continue;
      if tv<v[0] and i+k<=m:
        tv=v[0]
        tk=k
    ans+=tv
    if tk: dic[tk].popleft();
    
    
print(ans)