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 (2.11.7) (judge verdict: Wrong Answer)


import scala.collection.mutable.ArrayBuffer
import scala.io


object Main {
  def main(args: Array[String]) = {
    val input = for (line <- io.Source.stdin.getLines) yield {
      val numAry = line.split(' ').map(_.toInt)
      numAry
    }

    val NM = input.next()
    val N = NM(0)
    val M = NM(1)

    val ABarr = ArrayBuffer.empty[Array[Int]]
    while (input.hasNext) {
      ABarr.append(input.next)
    }

    var count = 0
    var sumPrice = 0
    val sortedArr = ABarr.map {
      elem => (elem(0), elem(1))
    }.sortBy {
      case (price, num) => -1 * price
    }
      
    var i = 0
    while(count > M){
      val (price, num) = sortedArr(i)

      if(count + num > M){
        println(sumPrice + price * (M - count))
        count += num
      }else{
        sumPrice += sumPrice
        count += num
      }
    }

  }
}