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 (3.4.3) (judge verdict: Runtime Error)

import sys
import bisect
import numpy as np

if sys.platform =='ios':
	sys.stdin=open('Untitled.txt')
input = sys.stdin.readline
def INT(): return int(input())
def MAP(): return [int(s) for s in input().split()]

N, M = MAP()
A = [MAP() for _ in range(N)]

A.sort()
A_day = [a[0] for a in A]
A_salary = [a[1] for a in A]

salary = 0
for i in range(M):
	print(A_day, M-i+1)
	idx = bisect.bisect_left(A_day, M-i+1)
	if idx == 0: continue
	todo = np.argmax(A_salary[:idx])
	print(i, idx, todo, A[todo], A_day[todo], A_salary[todo])
	salary += A_salary[todo]
	A_day.pop(todo)
	A_salary.pop(todo)

print(salary)