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: Wrong Answer)

object Main extends App {
  import scala.io.StdIn._
  val Array(n, m) = readLine().split(" ").map(_.toInt)
  val a = new Array[Long](n)
  val map = scala.collection.mutable.HashMap[Long,Long]()
  for(i <- (0 until(n))) {
    val Array(aa, bb) = readLine().split(" ").map(_.toInt)
    a(i) = aa
    map(aa) = if(map.contains(aa)) map(aa) + bb else bb
  }
  val aA = a.sorted
  var ans = 0L
  var ind = 0
  for(i <- (1 to m)){
    if(map(aA(ind)) == 0){
      ind += 1
    }
    ans += aA(ind)
    map(aA(ind)) -= 1
  }
  println(ans)
}