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.util.*;

public class Main {
  
  public static class Drink {
    public int A;
    public int B;
    public Drink(int a, int b) {
      A = a;
      B = b;
    }
  }
  
  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int N, M;
    N = sc.nextInt();
    M = sc.nextInt();
    
    long totalYen = 0L;
    List<Drink> list = new ArrayList<>();
    for (int i=0; i<N; i++) {
      list.add(new Drink(sc.nextInt(), sc.nextInt()));
    }
    Collections.sort(list, new Comparator<Drink>() {
      @Override
      public int compare(Drink d1, Drink d2) {
        return d1.A-d2.A;
      }
    });
    
    for (int i=0; i<list.size(); i++) {
      Drink d = list.get(i);
      long yen = d.A*d.B;
      if (M - d.B > 0) {
        totalYen += yen;
        M -= d.B;
      } else if (M - d.B <= 0) {
        for (int j=1; j<=d.B && M>=0; j++, M--) {
          totalYen += d.A;
        }
      }
      if (M == 0) {
        System.out.println(totalYen);
        return;
      } 
    }
    return;
  }
}