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) {
		// TODO 自動生成されたメソッド・スタブ

		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int m = sc.nextInt();
		long[] a = new long[n];
		long[] b = new long[n];
		int count = 0;
		long sum = 0;

		Map<Long, Long> map = new TreeMap<>();
		for (int i = 0; i < n; i++) {
			a[i] = sc.nextInt();
			b[i] = sc.nextInt();
			if (map.containsKey((a[i]))) {
				map.put((a[i]), map.get((a[i])) + b[i]);
			} else {
				map.put((a[i]), b[i]);
			}
		}

		for (Map.Entry<Long, Long> me : map.entrySet()) {
			if (me.getValue() < m - count) {
				sum += me.getKey() * me.getValue();
				count += me.getValue();
			} else {
				sum += me.getKey() * (m - count);
				System.out.println(sum);
				break;
			}






		}

	}
}