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 Kotlin (1.0.0) (judge verdict: Time Limit Exceeded)

fun main(args: Array<String>) {
    val (n, m) = readLine()!!.split(" ").map { it.toInt() }
    val abList = mutableListOf<Pair<Int, Int>>()
    (1..n).forEach {
        val (a, b) = readLine()!!.split(" ").map { it.toInt() }
        abList.add(Pair(a, b))
    }

    abList.sortBy { pair -> pair.first }

    var sum = 0
    for (i in m-1 downTo 0) {
        val remainDays = m - i
        val workableDays = abList.filter { pair -> pair.first <= remainDays }
        val maxReword = workableDays.maxBy { pair -> pair.second }
        if (maxReword != null ) run { sum += maxReword.second }
        abList.remove(maxReword)
    }

    println(sum)

}