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 ccc
implicit none

integer(8) :: n, m, i, j, res
integer(8),allocatable,dimension(:) :: a, b

read*, n, m
allocate(a(n),b(n))

do i=1,n
read*, a(i), b(i)
end do

call heapsortkai(n,a,b)

j=0
res=0
do while(m>0)
j=j+1
m=m-b(j)
res=res+a(j)*b(j)
end do

res=res-abs(m)*a(j)

write(*,'(i0)') res

end program

subroutine heapsortkai(n,array1,array2)
  implicit none
  integer(8),intent(in) :: n
  integer(8),intent(inout) :: array1(1:n), array2(1:n)
 
  integer(8) ::i,k,j,l
  integer(8) :: t1, t2
 
  if(n.le.0)then
     write(6,*)"Error, at heapsort"; stop
  endif
  if(n.eq.1)return

  l=n/2+1
  k=n
  do while(k.ne.1)
     if(l.gt.1)then
        l=l-1
        t1=array1(l)
        t2=array2(l)
     else
        t1=array1(k)
        t2=array2(k)
        array1(k)=array1(1)
        array2(k)=array2(1)
        k=k-1
        if(k.eq.1) then
           array1(1)=t1
           array2(1)=t2
           exit
        endif
     endif
     i=l
     j=l+l
     do while(j.le.k)
        if(j.lt.k)then
           if(array1(j).lt.array1(j+1))j=j+1
        endif
        if (t1.lt.array1(j))then
           array1(i)=array1(j)
           array2(i)=array2(j)
           i=j
           j=j+j
        else
           j=k+1
        endif
     enddo
     array1(i)=t1
     array2(i)=t2
  enddo

  return
end subroutine heapsortkai