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: Accepted)

import java.util.*;

public class Main{
    public static void main(String[] args) {
        Scanner scan = new Scanner(System.in);
        int N = scan.nextInt();
        int M = scan.nextInt();

        Integer[] A = new Integer[N];
        Integer[] B = new Integer[N];

        for(int i = 0; i < N; i++){
            A[i] = scan.nextInt();
            B[i] = scan.nextInt();
        }

        Integer[] idx = new Integer[N];
        Arrays.setAll(idx, i -> i);
        Arrays.sort(idx, Comparator.comparingInt(i -> A[i]));

        long count = 0, sum = 0;
        for(int i = 0; i < N; i++){
            count += B[idx[i]];
            sum += (long)A[idx[i]] * B[idx[i]];
            if(count > M){
                sum -= (long)(count - M) * A[idx[i]];
                break;
            }
        }

        System.out.println(sum);
    }
}