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)

object Main {

  def main(args: Array[String]): Unit = {
    val sc = new java.util.Scanner(System.in)
    val N, M = sc.nextInt()
    val ABs = Array.fill(N, 2)(sc.nextInt())
    val s = ABs.sortBy(_.head)
    println(s.foldLeft((0, M)) {
      case ((acc, m), arr) if m == 0 =>
        (acc, 0)
      case ((acc, m), arr) =>
        val a = arr.head
        val b = arr.last
        val buy = Math.min(m, b)
        (acc + (a * buy), m - buy)
    }._1)
  }
}