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 Haskell (GHC 7.10.3) (judge verdict: Accepted)

{-# LANGUAGE BangPatterns #-}

import Control.Monad
import Data.List
import Data.Maybe
import qualified Data.ByteString.Char8 as C8

type Pair = (Int, Int)

main :: IO ()
main = do
  [n,m] <- readInts
  ps <- sortOn fst <$> replicateM n readPair
  print $ solve m ps
 
solve :: Int -> [Pair] -> Int
solve m = fst . foldl' f e
  where
    e = (0,0)
    f (!s,!n) (a,b)
      | n == m = (s,n)
      | c > m = (s+a*(m-n),m) -- until ((m==).snd) (\(!s,!n) -> (s+a,n+1)) (s,n)
      | otherwise = (s+a*b, c)
      where
        c = n+b

-- utils
readInts :: IO [Int]
readInts = map (fst . fromJust . C8.readInt) . C8.words <$> C8.getLine

readPair :: IO Pair
readPair = do
  [x,y] <- readInts
  return (x,y)