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();
		int M = sc.nextInt();
		long money = 0;

		int[][] ary = new int[2][N];

		for (int i = 0; i < N; i++) {
			ary[0][i] = sc.nextInt();
			ary[1][i] = sc.nextInt();
		}

		while (M > 0) {
			int midx = minIdx(ary);
			if (ary[1][midx] > M) {
				money += (long) ary[0][midx] * M;
				M = 0;
			} else {
				money += (long) ary[0][midx] * ary[1][midx];
				ary[0][midx] = 1000000000;
				M -= ary[1][midx];
			}
		}

		System.out.println(money);
	}

	static int minIdx(int[][] a) {
		int idx = 0;
		int min = a[0][0];
		for (int i = 0; i < a[0].length; i++) {
			if (a[0][i] < min) {
				idx = i;
				min = a[0][i];
			}
		}
		return idx;
	}
}