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: Runtime Error)

class Work:
    count=0
    def __init__(self,payday,money):
        Work.count+=1
        self.count=count
        self.payday=payday
        self.money=money
 #moneyが大きい方、同じなら早くもらえる方、それも同じなら最初に書いてある方が強い方
    def __gt__(self,other):
        if self.money>other.money:
            return True
        if self.money==other.money:
            if self.payday<other.payday:
                return True
            if self.payday==other.payday:
                if self.count<other.count:
                    return True
        return False

    def __ft__(self,other):
        if self.money<other.money:
            return True
        if self.money==other.money:
            if self.payday>other.payday:
                return True
            if self.payday==other.payday:
                if self.count>other.count:
                    return True
        return False

#     def __eq__(self,other):
#         if self.money==other.money and self.payday==other.payday:
#             return True
#         return False
#     def __fe__(self,other):
#         return self.__ft__(other) or self.__eq__(other)
#     def __ge__(self,other):
#         return self.__gt__(other) or self.__eq__(other)

#バイトの数、いつまでに稼ぐのか 
n,limit=[int(i) for i in input().split()]

#バイトのリストを作り、クラスを入れる O(n)
baitolist=[]
for i in range(n):
    payday,money=[int(i) for i in input().split()]
    work=Work(payday,money)
    baitolist.append(work)

#日数でランク付け　O(n)
summoney=0    
rank=[[] for i in range(limit+1)]
for baito in baitolist:
    if baito.payday<=limit:
        rank[baito.payday].append(baito)

accessible=[]
count=1
while baitolist and count<=limit:#最悪O(limit)
    for baito in rank[count]:
        accessible.append(baito)
    accessible.sort()
    if accessible!=[]:
        dowork=accessible.pop()
        summoney+=dowork.money
    count+=1
    
    
print(summoney)