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: Time Limit Exceeded)

import java.util.Scanner;
public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(), m = sc.nextInt();
		int [] price = new int [n];
		int [] number = new int [n];
		for(int i = 0; i < n; i++) {
			price[i] = sc.nextInt();
			number[i] = sc.nextInt();
		}
		int temp = 0;
		for(int i = 0; i < n; i++)
			for(int k = n-1; k > i; k--) {
				if(price[k-1] > price[k]) {
					temp = price[k-1];
					price[k-1] = price[k];
					price[k] = temp;
					
					temp = number[k-1];
					number[k-1] = number[k];
					number[k] = temp;
				}
			}
		long sum = 0;
		int i = 0;
		while(true) {
			if(number[i] >= m) {
				sum = sum + (long)price[i] * m;
				m = m - number[i];
				System.out.println(sum);
				break;
			}else if(number[i] < m) {
				sum = sum + (long)price[i] * number[i];
				m = m - number[i];
			}
			i++;
		}
		sc.close();
	}

}