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: Accepted)

class Pair
  attr_accessor :first,:second
end

def make_pair(x,y)
  temp = Pair.new
  temp.first = x
  temp.second = y
  return temp
end

n, m = gets.chomp.split.map(&:to_i)
s = []
n.times do |i|
  a, b = gets.chomp.split.map(&:to_i)
  s.push(make_pair(a, b))
end
s = s.sort{|i, j| i.first <=> j.first}
c = 0
n.times do |i|
  if m >= s[i].second
    m -= s[i].second
    c += s[i].first * s[i].second
  else
    s[i].second.times do |j|
      if m > 0
        c += s[i].first
        m -= 1
      else
        break
      end
    end
  end
end
p c
exit