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.ArrayList;
import java.util.List;
import java.util.Scanner;

public class Main {
	public static long count;
	public static int m;
	public static long[] a;
	public static int[] b;
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int n = sc.nextInt();
		m = sc.nextInt();
		
		
		a = new long[n];
		b = new int[n];
		
		for (int i = 0; i < n; i++) {
			a[i] = sc.nextLong();
			b[i] = sc.nextInt();
		}
		
		long[][] c = new long[n][2];
		
		List<Integer> used = new ArrayList<>();
		
		for (int i = 0; i < a.length; i++) {
			int index = 0;
			long min = Long.MAX_VALUE;
			for (int j = 0; j < a.length; j++) {
				if (!used.contains(j) && min > a[j]) {
					min = a[j];
					index = j;
				}
			}
			c[i][0] = a[index];
			c[i][1] = b[index];
			used.add(index);
		}
		
		used = null;
		
		count = 0;
		long money = 0L;
		
		int tmp = 0;
		
		while (count < m) {
			for (int i = tmp; i < c.length; i++) {
				if (c[i][1] > 0) {
					long num = Math.min(m - count, c[i][1]);
					count += num;
					money += c[i][0] * num;
					c[i][1] -= num;
					break;
				} else {
					++tmp;
				}
			}
		}
		
		System.out.println(money);
	}
}