Problem:
Score : 300 points

Problem Statement

Hearing that energy drinks increase rating in those sites, Takahashi decides to buy up M cans of energy drinks.

There are N stores that sell energy drinks. In the i-th store, he can buy at most B_i cans of energy drinks for A_i yen (the currency of Japan) each.

What is the minimum amount of money with which he can buy M cans of energy drinks?

It is guaranteed that, in the given inputs, a sufficient amount of money can always buy M cans of energy drinks.

Constraints

All values in input are integers.

1 \leq N, M \leq 10^5

1 \leq A_i \leq 10^9

1 \leq B_i \leq 10^5

B_1 + ... + B_N \geq M

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 minimum amount of money with which Takahashi can buy M cans of energy drinks.

Sample Input 1

2 5
4 9
2 4

Sample Output 1

12

With 12 yen, we can buy one drink at the first store and four drinks at the second store, for the total of five drinks. However, we cannot buy 5 drinks with 11 yen or less.

Sample Input 2

4 30
6 18
2 5
3 10
7 9

Sample Output 2

130

Sample Input 3

1 100000
1000000000 100000

Sample Output 3

100000000000000

The output may not fit into a 32-bit integer type.

Solution in Scala (2.13.1) (judge verdict: Accepted)

object Main extends App {
  val inputs: Seq[String] = scala.io.Source.stdin.getLines().toList
  val nm = inputs.head.split(" ").map(_.toInt)
  val n  = nm(0)
  val m  = nm(1)

  println(solve(n,m, inputs.tail))

  def solve(n: Int, m: Int, ab: Seq[String]): Long = {
    val hash = scala.collection.mutable.HashMap.empty[Long, Long]
    var ans = 0L
    var nowStock = 0L
    val keySet = ab.map { line =>
      val lineNum = line.split(" ").map(_.toLong)
      val key = lineNum(0)
      hash.update(key, hash.applyOrElse(key, (n: Long) => 0L) + lineNum(1))
      key
    }.distinct.sorted

    keySet.foreach { price =>
      val stock = hash(price)

      if (m < stock + nowStock) {
        val v = (m - nowStock)
        ans = ans + v * price
        nowStock = nowStock + v
      } else {
        nowStock = nowStock + stock
        ans = ans + (stock * price)
      }

      if (m == nowStock) {
        return ans
      }

    }

    -1L
  }
}