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 (7.3.0) (judge verdict: Accepted)

from sys import stdin
import sys
import math
from functools import reduce
import functools
import itertools
from collections import deque,Counter,defaultdict
from operator import mul
import copy
# ! /usr/bin/env python
# -*- coding: utf-8 -*-
import heapq
sys.setrecursionlimit(10**6)
# INF =  float("inf")
INF = 10**18
import bisect
import statistics
mod = 10**9+7
# mod = 998244353

task = []
heapq.heapify(task)

N, M = map(int, input().split())

A = [[] for i in range(M+1)]
for i in range(N):
    a, b = map(int, input().split())
    if a > M: continue
    A[a].append(b)

ans = 0
for i in range(1,M+1):
    for b in A[i]:
        heapq.heappush(task, -b)
    if len(task) > 0:
        ans += heapq.heappop(task)*(-1)

print(ans)