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)

n,m = gets.chomp.split.map{|i|i = i.to_i}
a, b=[],[]
n.times do|i|
  arr = gets.chomp.split.map{|i|i = i.to_i}
	a << arr[0]
	b << arr[1]
end
#puts "a = #{a}"
#puts "b = #{b}"
#puts "b.index(b.min) = #{b.index(b.min)}"

drink_num=0
sum_money=0
while drink_num<m do
	i=a.index(a.min)
	if (m-drink_num) < b[i] then
		sum_money += (m - drink_num) * a[i]
		break
	else
		drink_num += b[i]
		sum_money += b[i] * a[i]
		a.delete_at(i)
		b.delete_at(i)
	end
	#puts "now b = #{b}"
end
#puts "drink_num=#{drink_num}, sum_money=#{sum_money}"
puts sum_money