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)

n,m = gets.split(" ").map(&:to_i)
arr = n.times.map{gets.split.map(&:to_i)}

hash = {}
res = 0
arr.each do |i|
    hash[i[0]] = [[],0] if !hash.key?(i[0])
    hash[i[0]][0].push(i[1])
end
hash.each do |key,val|
    val[1] = val[0].sort!.last
end

(1..m).to_a.each do |i|
    max = [0, 0] #day, val
    (1..[i,hash.keys.max].min).to_a.each do |j|
        max = [j,hash[j][1]] if hash.key?(j) && !hash[j][0].empty? && hash[j][1] > max[1]
    end
    if max[1] != 0
        res += hash[max[0]][0].pop 
        hash[max[0]][1] = hash[max[0]][0].last.to_i
    end
end

p res