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: Wrong Answer)

from sys import stdin
import math
import fractions
from collections import deque
from collections import Counter
import itertools
import heapq

readline = stdin.buffer.readline

INF = 10 ** 10

# M日後まで得られる報酬の合計の最大値
N, M = [int(x) for x in readline().split()]

#Ai日後にBiの金
AB = [tuple(map(int, readline().split())) for _ in range(N)]


# 金が貰えて時間がかかる順にソート
AB.sort(key=lambda x: (x[1], x[0]), reverse=True)
AB = deque(AB)

ans = 0

while(AB):
    A, B = AB.popleft()
    if A <= M:
        ans += B
        M -= 1

    if (M <= 0):
        break

print(ans)