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) {
        final Scanner sc = new Scanner(System.in);
        final long N = sc.nextLong();
        final long M = sc.nextLong();

        final long[] A = new long[(int)N];
        final long[] B = new long[(int)N];

        final PriorityQueue<Map.Entry<Long, Integer>> pq = new PriorityQueue<>((e1, e2) -> {
            if (e1.getKey() < e2.getKey()) {
                return -1;
            }
            if (e1.getKey() > e2.getKey()) {
                return +1;
            }
            return 0;
        });

        for (int i = 0; i < N; ++i) {
            A[i] = sc.nextLong();
            pq.offer(new AbstractMap.SimpleEntry<>(A[i], i));
            B[i] = sc.nextLong();
        }

        long leftToBuy = M;
        long money = 0;

        while (!pq.isEmpty() && leftToBuy > 0) {
            final Map.Entry<Long, Integer> entry = pq.poll();
            final long price = entry.getKey();
            final int index = entry.getValue();
            money += (Math.min(B[index], leftToBuy)) * price;
            leftToBuy -= Math.min(B[index], leftToBuy);
        }

        System.out.println(money);
        System.out.flush();
        sc.close();
    }
}