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.Scanner;

public class Main {
	public static void main(String args[]) {
		Scanner scn = new Scanner(System.in);
		int N = scn.nextInt();
		int M = scn.nextInt();
		Drink[] A = new Drink[N];
		for (int i = 0; i < N; i++) {
			A[i] = new Drink(scn.nextInt(), scn.nextInt());
		}
		Arrays.sort(A);
		long ans = 0;
		for(int i = 0;i < N;i++) {
			if(M > A[i].getMax()) {
				ans += A[i].getPrice() * A[i].getMax();
				M -= A[i].getMax();
			}else {
				ans += A[i].getPrice() * M;
				break;
			}
		}
		System.out.println(ans);
	}
}

class Drink implements Comparable<Drink>{
	long price, max;

	Drink(long price, int max) {
		this.price = price;
		this.max = max;
	}

	long getPrice() {
		return this.price;
	}

	long getMax() {
		return this.max;
	}

	@Override
	public int compareTo(Drink o) {
		if(this.getPrice() < o.getPrice()) {
			return -1;
		}else {
			if(this.getPrice() > o.getPrice()) {
				return 1;
			}else {
				return 0;
			}
		}
	}
}