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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main{

    public static void main(String[] args) {

        Scanner scan = new Scanner(System.in);

        int N = scan.nextInt();

        int M = scan.nextInt();


        long A[] = new long[N];

        long B[] = new long[N];

        Map<Long,Long> AB = new HashMap<Long,Long>();

        for(int i = 0; i < N; i++){

            A[i] = scan.nextInt();
            B[i] = scan.nextInt();

            AB.put(A[i],B[i]);

        }

        long sortList[] = new long[N];

        sortList = A;

        Arrays.sort(sortList);

        //今いくつ買った？
        long buyCnt = 0;

        //いくら払った？
        long pay = 0;

        int cnt = 0;

        while(true){

            //買わねばならない数
            long mustBuy = M - buyCnt;

            if(mustBuy == 0){
                break;
            }

            //今買う数
            long nowBuy = AB.get(sortList[cnt]);

            if(nowBuy > mustBuy){
                nowBuy = mustBuy;
            }

            pay = pay + nowBuy * sortList[cnt];

            AB.remove(sortList[cnt]);

            buyCnt = buyCnt + nowBuy;
            cnt++;

        }

        System.out.print(pay);

    }
}