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.Collections;
import java.util.List;
import java.util.Objects;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		int m = in.nextInt();
		ArrayList<P> list = new ArrayList<>();
		for (int i = 0; i < n; i++) {
			list.add(new P(in.nextLong(), in.nextInt()));
		}
		Collections.sort(list);
		long ret = 0;
		for (P p : list) {
			if (m > p.b) {
				m -= p.b;
				ret += p.a * p.b;
			} else {
				ret += p.a * m;
				break;
			}
		}
		System.out.println(ret);
	}

	static class P implements Comparable<P> {
		long a;
		int b;

		P(long a , int b) {
			this.a = a;
			this.b = b;
		}

		@Override
		public int compareTo(P o) {
			long d = this.a - o.a;
			if (d > 0) {
				return 1;
			} else if (d == 0) {
				return 0;
			}  else {
				return -1;
			}
		}
	}
}