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)::n,m,i,p=0,q=0
	integer(8),allocatable::a(:),b(:)
	
	read*,n,m
	allocate(a(n),b(n))
	
	do i =1,n
		read*,a(i),b(i)
	enddo
	
	call h2(a,b,n)
	
	do i =1,n
		if(m-q <= b(i))then
			p = p + a(i)*(m-q)
			exit
		endif
		q = q+b(i)
		p = p+a(i)*b(i)
	enddo
	
	print"(i0)",p
end program main

subroutine h2(a,b,n)
implicit None
	integer(8)::a(n),b(n),x,y
	integer(8)::n,i,ie,i0,i1,i2
	
	do i = n/2,1,-1
		x = a(i);y = b(i);ie = n + 1; i0 = i; i1 = i0 * 2
		do while(i1 < ie)
			i2 = i1 + 1
			if(i2 < ie .and. a(i1) < a(i2)) i1 = i2
			if(a(i1) > x)then
				a(i0) = a(i1);a(i1) = x
				b(i0) = b(i1);b(i1) = y
				else
				exit
			endif
			i0 = i1;i1 = i0 *2
		enddo
	enddo
	do ie = n,2,-1
		x = a(ie);y = b(ie)
		a(ie) = a(1);b(ie) = b(1)
		a(1) = x;b(1) = y
		i0 =1;i1 =2
		do while(i1 <ie)
			i2 = i1 +1
			if(i2 < ie .and. a(i1) < a(i2)) i1 = i2
			if(a(i1) > x)then
				a(i0) = a(i1);a(i1) = x
				b(i0) = b(i1);b(i1) = y
				else
				exit
			endif
			i0 = i1;i1 = i0 *2
		enddo
	enddo
end subroutine h2