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 OCaml (4.02.3) (judge verdict: Accepted)

let input () =
let (n,m) = Scanf.sscanf (read_line ()) "%d %d" (fun a b -> (a,b)) in
let rec loop count list =
  match count with
    0 -> list
  | x -> let (a,b) = Scanf.sscanf (read_line ()) "%d %d" (fun a b -> (a,b)) in
          loop (count - 1) ((a,b)::list)
in (m,loop n []);;

let solve () =
  let (m, list) = input() in
  let sort_list = List.sort (fun (a,_) (b,_) -> if a > b then 1 else -1) list in
  let rec loop ans m l =
    let (a,b) = List.hd l in
    if m > b then
      loop (ans + (a*b)) (m-b) (List.tl l)
    else print_int (ans + m*a)
  in loop 0 m sort_list;;

solve();;