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: Accepted)

import sys
from collections import Counter, deque, defaultdict
from math import factorial
from fractions import Fraction
import heapq, bisect
import math
import itertools
sys.setrecursionlimit(10 ** 5 + 10)
INF = 10**15 +5
def input(): return sys.stdin.readline().strip()
def INT(): return int(input())
def MAP(): return map(int, input().split())
def LIST(): return list(map(int, input().split()))

n, m = MAP()
bef = []
for i in range(n):
    a, b = MAP()
    bef.append([a, b])
candi = []
res = 0
heapq.heapify(bef)
heapq.heapify(candi)

for i in range(m):
    while True:
        if len(bef) == 0:
            break
        temp = heapq.heappop(bef)
        if temp[0] > i+1:
            heapq.heappush(bef, temp)
            break
        heapq.heappush(candi, temp[1] * -1)
    if len(candi) != 0:
        res += heapq.heappop(candi)*-1

print(res)