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)

import com.sun.org.apache.xpath.internal.operations.Bool
import java.lang.Math.*

const val MOD = 1000000007L

fun main(args: Array<String>) {
    val (n, m) = readIntList()

    val ab = Array(m + 1) { index ->
        return@Array mutableListOf<Int>()
    }

    n.rep {
        val (a, b) = readIntList()
        if (a <= m) {
            ab[a].add(b)
        }
    }

    var ans = 0
    (m + 1).rep(1) {
        val abMax = ab[it].max() ?: 0
        ans += abMax
        if (abMax != 0) {
            ab[it].remove(abMax)
            if (it + 1 < ab.size) {
                ab[it + 1].addAll(ab[it])
            }
        }
    }

    println(ans)
}

fun readList() = readLine()!!.split(' ')
fun readInt() = readLine()!!.split(' ').map(String::toInt).first()
fun readIntList() = readList().map(String::toInt)
fun readLongList() = readList().map(String::toLong)
fun readDoubleList() = readList().map(String::toDouble)

fun readPair(): Pair<String, String> {
    val splite = readLine()!!.split(' ')
    return Pair(splite[0], splite[1])
}

fun <T> List<T>.copy() = this.map { it }

inline fun <reified T> dualArray(h: Int, w: Int, init: (Int, Int) -> T) {
    Array(h) { i ->
        return@Array Array(w) { j ->
            return@Array init(i, j)
        }
    }
}

// 速度のためforEachは避ける
inline fun Int.rep(start: Int = 0, shiftLast: Boolean = false, body: (Int) -> Unit) {
    for (i in start until if (shiftLast) this + start else this) {
        body(i)
    }
}

inline fun Long.rep(start: Long = 0, shiftLast: Boolean = false, body: (Long) -> Unit) {
    for (i in start until if (shiftLast) this + start else this) {
        body(i)
    }
}

// 約数のList
fun divisor(value: Long): List<Long> {
    val max = sqrt(value.toDouble()).toLong()
    val former = mutableListOf<Long>()
    val latter = mutableListOf<Long>()
    (1..max).filter { value % it == 0L }
        .forEach {
            former.add(it)
            latter.add(value / it)
        }
    former.addAll(latter.reversed())
    return former
}

// 範囲内の素数を取得
// fromだけ指定すると戻り値の個数で素数判定ができる
fun prime(from: Long, to: Long = from): List<Long> {
    return (from..to).filter { i ->
        val max = Math.sqrt(i.toDouble()).toLong()
        (2..max).all { j -> i % j != 0L }
    }
}

// 素因数分解
fun decom(value: Long): List<Long> {
    if (value == 1L) return listOf(1)
    val max = Math.sqrt(value.toDouble()).toLong()
    return prime(2, max).filter { value % it == 0L }
}

// 最大公約数
fun gcd(a: Long, b: Long): Long {
    return if (a % b == 0L) b else gcd(b, a % b)
}

// 文字列を入れ替え
fun swap(base: String, a: String, b: String): String {
    return base.map {
        when (it) {
            a.toCharArray()[0] -> b
            b.toCharArray()[0] -> a
            else -> it.toString()
        }
    }.joinToString()
}

// 二分探索
fun List<Long>.binarySearch(target: Long, _start: Int = 0, _end: Int = this.size): Int {
    var start = (_start - 1).toLong()
    var end = _end.toLong()
    while (end - start > 1) {
        val mid = (start + end) / 2
        if (this[mid.toInt()] < target) {
            start = mid
        } else {
            end = mid
        }
    }
    return end.toInt()
}

// bit全探索
inline fun Int.bitAllSearch(body: (Array<Int>) -> Unit) {
    for (bit in 0 until (1 shl this)) {
        val s = mutableListOf<Int>()
        for (i in 0 until this) {
            if ((bit and (1 shl i)) shr i == 1) { // i が bit に入るかどうか
                s.add(i)
            }
        }
        body(s.toTypedArray())
    }
}