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: Wrong Answer)

import java.math.BigDecimal;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {

    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int N = Integer.parseInt(sc.next());
        int M = Integer.parseInt(sc.next());
        int[] Ai = new int[N];
        int[] Bi = new int[N];
        Map<Integer, Integer> map = Collections.synchronizedMap(new HashMap<>());

        //AiとBiの値を配列に格納、AiとBiの値の対応付けをする
        for (int i = 0; i < N; i++) {

            Ai[i] = Integer.parseInt(sc.next());
            Bi[i] = Integer.parseInt(sc.next());

            if(map.containsKey(Ai[i])) {

            	map.put(Ai[i], map.get(Ai[i]) + Bi[i]);
            } else {

            map.put(Ai[i], Bi[i]);
            }
        }

        sc.close();

        //Aiを昇順にソート
        Arrays.sort(Ai);

        System.out.println(func(M, Ai, map, BigDecimal.ZERO, BigDecimal.ZERO, 0));

    }

    //桁あふれ対策でBigDecimalを使って計算する
    public static BigDecimal func(int M, int[] Ai, Map<Integer, Integer> map, BigDecimal sum, BigDecimal pay, int i) {

        //栄養ドリンクが安い店から順に欲しい本数になるまで在庫分購入する
        switch (sum.compareTo(BigDecimal.valueOf(M))) {

        case -1:
            sum = sum.add(BigDecimal.valueOf(map.get(Ai[i])));
            pay = pay.add(BigDecimal.valueOf(map.get(Ai[i])).multiply(BigDecimal.valueOf(Ai[i])));
            i++;
            return func(M, Ai, map, sum, pay, i);

        case 0:

            return pay;

        //買いすぎた本数分払い戻しして支払い総額を求める
        default :
            pay = pay.subtract(((sum.subtract(BigDecimal.valueOf(M))).multiply(BigDecimal.valueOf(Ai[i-1]))));
            return pay;
        }

    }
}