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 Fortran (GNU Fortran 9.2.1) (judge verdict: Time Limit Exceeded)

program answer
  implicit none
  integer(16) :: i, N, M, sum, count, last, aa
  integer(16), allocatable :: A(:), B(:)
  read(*,*) N, M
  allocate(A(N),B(N))
  do i = 1, N
     read(*,*) A(i), B(i)
  end do

  sum=0
  count=0
  do
     if(count>=M) then
        exit
     end if
     aa=dot_product(B(minloc(A)),B(minloc(A)))
     aa=int(sqrt(dble(aa)))
     sum=sum+minval(A)*aa
     count=count+aa
     last=minval(A)
     A(minloc(A))=1000000001
  end do

  if(count>M) then
     sum=sum-(count-M)*last
  end if

  write(*,*) sum
  deallocate(A,B)
  stop
 end program answer