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.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {
    
    static class Pair {
        int left;
        int right;

        Pair(int l, int r) {
            left = l;
            right = r;
        }
    }
    public static void main(String[] args) {
        Scanner std = new Scanner(System.in);
        int n = std.nextInt();
        int m = std.nextInt();
        int[] as = new int[n];
        int[] bs = new int[n];
        long[][] dp = new long[n + 1][m + 1];
        PriorityQueue<Pair> pq = new PriorityQueue<>(new Comparator<Pair>() {
            @Override
            public int compare(Pair o1, Pair o2) {
                return o1.right - o2.right;
            }
        });
        for (int i = 0; i < n; i++) {
            as[i] = std.nextInt();
            bs[i] = std.nextInt();
            pq.add(new Pair(bs[i], as[i]));
        }

        for (int i = 0; i < dp.length; i++) {
            Arrays.fill(dp[i], Long.MAX_VALUE);
        }

        int count = 0;
        long ans = 0;
        while (count != m || !pq.isEmpty()) {
            Pair p = pq.poll();
            int c = Math.min(m - count, p.left);
            count += c;
            ans += p.right * c;
        }

        System.out.println(ans);

    }
}