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: Time Limit Exceeded)

Program ABC121C

integer(8) n,m,c,p,x
integer(8),allocatable::a(:),b(:)

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

c=0
p=0

do i=1,n
 do j=i+1,n
  if (a(i)>a(j)) then
   x=a(i)
   a(i)=a(j)
   a(j)=x
   x=b(i)
   b(i)=b(j)
   b(j)=x
  end if
 end do
end do

do i=1,n
 c=c+a(i)*b(i)
 p=p+b(i)
 if (p>=m) then
  c=c-a(i)*(p-m)
  go to 10
 end if
end do
 
10 write(*,*) c

stop
end