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 Java7 (OpenJDK 1.7.0) (judge verdict: Time Limit Exceeded)



import java.util.Arrays;
import java.util.HashMap;
import java.util.Scanner;

public class Main{
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		long[] a = new long[n];
		long[] b = new long[n];
		HashMap<Long, Long> map = new HashMap<>();

		for (int i = 0; i < n; i++) {
			a[i] = sc.nextLong();
			b[i] = sc.nextLong();
		}

		long[] ac = a.clone();

		Arrays.sort(ac);

		long ans = 0;
		long sum = 0;

		out_of_loop:
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				if (ac[i] == a[j]) {

					if (sum + b[j] >= m) {
						ans += ac[i] * (m - sum);
						break out_of_loop;
					} else {
						ans += ac[i] * b[j];
						sum += b[j];
					}
					a[j] = -1;
				}
			}
		}
		System.out.println(ans);

	}

}