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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

nm = gets.chomp.split
n = nm[0].to_i
m = nm[1].to_i

a = []
b = []
n.times do
  ab = gets.chomp.split
  a << ab[0].to_i
  b << ab[1].to_i
end

money = 0
drinks = 0
while drinks < m
  store_index = a.index(a.min)
  price = a[store_index]
  drink = b[store_index]

  if drinks + drink <= m
    drinks += drink
    money += (drink * price)
  else
    money += ((m - drinks) * price)
    break
  end

  a.delete_at(store_index)
  b.delete_at(store_index)
end

puts money