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.Arrays;
import java.util.Comparator;
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();
		Drink[] d = new Drink[n];
		for (int i = 0; i < n; i++) {
			d[i] = new Drink(sc.nextInt(), sc.nextInt());
		}
		Arrays.sort(d, new Comparator<Drink>() {
			public int compare(Drink d1, Drink d2) {
				if(d1.price == d2.price) {
					return d2.num-d1.num;
				}else {
					return d1.price-d2.price;
				}
			}
		});
		long sum = 0;
		for(Drink d1:d) {
			if(m>d1.num) {
				sum += (long)d1.num*(long)d1.price;
				m -= d1.num;
			}else {
				sum += (long)d1.price*(long)m;
				break;
			}
		}
		System.out.println(sum);
		sc.close();
	}
	public static class Drink {
		int price; int num;
		Drink(int a, int b){
			this.price = a;
			this.num = b;
		}
	}
}