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.Map;
import java.util.Scanner;
import java.util.TreeMap;
public class Main{

	public static void main(String[] args) {

		Scanner scan = new Scanner(System.in);

		int store = scan.nextInt();
		int drink = scan.nextInt();
		Map<Long,Integer> sansyo = new TreeMap<Long,Integer>();

		for(int i=0;i<store;i++) {
			long money = scan.nextLong();
			int honsu = scan.nextInt();
			if(sansyo.containsKey(money)) {
				sansyo.put(money,sansyo.get(money)+honsu);
			}else {
			sansyo.put(money, honsu);
		}
		}

		long sum = 0l;
		loop:for(Long x:sansyo.keySet()) {

			if(sansyo.get(x) >=drink) {
				sum += drink*x;
				break loop;
			}else {
				sum += sansyo.get(x)*x;
				drink -= sansyo.get(x);
			}
		}
		System.out.println(sum);
		scan.close();

	}

}