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 Kotlin (1.0.0) (judge verdict: Accepted)

import java.util.*

fun main(args: Array<String>) {
  val sc = Scanner(System.`in`)
  val n = sc.nextLong()
  val m = sc.nextLong()
  val ab = (1..n).map { Pair(sc.nextLong(), sc.nextLong()) }

  val map = mutableMapOf<Long, Long>()
  for (abi in ab) {
    val a = abi.first
    val b = abi.second
    val tmp = map.getOrElse(a) { 0 }
    map.put(a, b + tmp)
  }

  val keys = map.keys.sorted()
  var count = 0L
  var ans = 0L
  for (key in keys) {
    val value = map.getOrElse(key) { 0 }
    if (count + value >= m) {
      val tmp = count + value - m
      val cost = key * (value - tmp)
      ans += cost
      break
    }
    count += value
    ans += value * key
  }

  println(ans)
}