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 Kotlin (1.0.0) (judge verdict: Accepted)

fun main(args: Array<String>) {
    val (n, m) = readLine()!!.split(" ").map { it.toLong() }
    val storeList = (1..n).map { readLine()!!.split(" ").map { it.toLong() } }

    val sortedStore = storeList.sortedBy { it.first()}

    // お店の最大本数よりも購入したい本数が
    // -> 少ない場合 購入したい本数*価格 を合計に足す
    // -> 多い場合 最大本数*価格を合計に足す 購入したい本数から最大本数を引く
    var wanted = m
    var total:Long = 0
    for(store in sortedStore){
        if(wanted < store[1]){
            total += store[0] * wanted
            break
        }else{
            total += store[0] * store[1]
            wanted -= store[1]
        }
    }

    println(total)
}