Problem: 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.

Examples

Input

2 5
4 9
2 4


Output

12


Input

4 30
6 18
2 5
3 10
7 9


Output

130


Input

1 100000
1000000000 100000


Output

100000000000000

Solution (in JAVA):

Solution:
import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.stream.Collectors;

public class Main {

	public static void main(String[] args) {
		try (Scanner scanner = new Scanner(System.in)) {
			int n = scanner.nextInt();
			int m = scanner.nextInt();
			scanner.nextLine();
			List<Store> list = new ArrayList<>();
			for (int i = 0; i < n; i++) {
				Store store = new Store();
				store.price = scanner.nextInt();
				store.amount = scanner.nextInt();
				list.add(store);
				scanner.nextLine();
			}
			List<Store> newList = list.stream().sorted((x, y) -> x.price - y.price).collect(Collectors.toList());
			long sum = 0;
			for (int i = 0; i < newList.size(); i++) {
				Store store = newList.get(i);
				if (store.amount >= m) {
					sum += (long) store.price * m;
					break;
				} else {
					sum += (long) store.price * store.amount;
					m -= store.amount;
				}
			}
			System.out.println(sum);
		}
	}

	private static class Store {
		int price;
		int amount;
	}
}