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 (gfortran v4.8.4) (judge verdict: Accepted)

program main
implicit none
integer(8) :: i, j, k
integer(8) :: n, m
integer(8), allocatable :: a(:), b(:)

read(*,*) n, m
allocate( a(n), b(n) )
do i = 1, n
   read(*,*) a(i), b(i)
end do
call heapsort2( n, a, b )

i = 0
j = 1
k = 0
do while( i+b(j) .le. m )
   k = k + a(j)*b(j)
   i = i + b(j)
   j = j + 1
end do
k = k + (m-i)*a(j)
write(*,'(i0)') k
      
stop
contains
subroutine heapsort2(n,array,array2)
  implicit none
  integer(8),intent(in) :: n
  integer(8),intent(inout) :: array(1:n), array2(1:n)
  integer(8) ::i,k,j,l
  integer(8) :: t, u

  l=n/2+1
  k=n
  do while(k.ne.1)
     if(l.gt.1)then
        l=l-1
        t=array(L)
        u=array2(L)
     else
        t=array(k)
        u=array2(k)
        array(k)=array(1)
        array2(k)=array2(1)
        k=k-1
        if(k.eq.1) then
           array(1)=t
           array2(1)=u
           exit
        end if
     end if
     i=l
     j=l+l
     do while(j.le.k)
        if(j.lt.k)then
           if(array(j).lt.array(j+1))j=j+1
        end if
        if (t.lt.array(j))then
           array(i)=array(j)
           array2(i)=array2(j)
           i=j
           j=j+j
        else
           j=k+1
        end if
     end do
     array(i)=t
     array2(i)=u
  end do
end subroutine heapsort2


end program main