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: Time Limit Exceeded)

import java.util.Scanner


object Main {
  val sc = new Scanner(System.in)

  def main(args: Array[String]): Unit = {
    val n = sc.nextInt
    val m = sc.nextInt

    val ab = Vector.fill(n)( (sc.nextLong, sc.nextLong) ).sorted
        .map{case (x,y) => (x,y,x*y) }
        .foldLeft(Vector.empty[(Long, Long, Long, Long)]){ case ( res, acc) =>
          res :+ (acc._1, acc._2, res.map(_._2).sum + acc._2, res.lastOption.getOrElse((0L, 0L, 0L, 0L))._4 + acc._3)
        }

    val dataTuple = ab.find(_._3 >= m).get

    val res = dataTuple._4 - dataTuple._1 * ( dataTuple._3 - m)

//    println(ab)
//    println(dataTuple)
    //    val r = ab.foldLeft(0L){()
    println(res)
  }
}