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: Wrong Answer)

import java.util.*

fun main(args:Array<String>){
    val scn= Scanner(System.`in`)
    val n=scn.nextInt()
    val m=scn.nextInt()
    var b=LongArray(n)
    var a=LongArray(n)
    var buy=0L
    var sum=0L

    for (i in 0 until n){
        a[i]=scn.nextLong()
        b[i]=scn.nextLong()
    }
    while (true){
        var idx=minIndex(a)
        sum+=a[idx]*b[idx]
        buy+=b[idx]
        if (buy>m) break
        a[idx]= Long.MAX_VALUE
    }
    sum-=(buy-m)*a[minIndex(a)]

    println(sum)
}

fun minIndex(x:LongArray):Int{
    var i=0
    var p=x[0]
    for (j in 1 until x.size){
        if (p>x[j]){
            i=j
        }
    }
    return i
}