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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

#!/usr/bin/ruby

n, m = gets.split.map { |s| s.to_i }
ab = []
(0...n).each { |i|
  ab[i] = gets.split.map { |s| s.to_i }
}
ab.sort! { |ab1, ab2|
  a1, b1 = ab1
  a2, b2 = ab2
  if b1 != b2
    b2 <=> b1
  else
    a1 <=> a2
  end
}
payments = (0..m).to_a.map { 0 }
count = 0
(0...ab.length).each { |i|
  a, b = ab.shift
  while a <= m && payments[a] > 0
    a += 1
  end
  next if a > m
  payments[a] = b
  count += 1
  break if count == m
  while payments[m] > 0
    m -= 1
  end
}
puts payments.reduce(:+)