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
import java.util.*;
import java.io.*;

class Shop implements Comparable<Shop>{
    long price;
    int stock;
    public Shop(long price, int stock){
        this.price = price;
        this.stock = stock;
    }

    public int compareTo(Shop o) {
        return Long.compare(this.price, o.price);
    }

    public String toString(){
        return "["+price+","+stock+"]";
    }
}

public class Main {
    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int M = sc.nextInt();
        LinkedList<Shop> shops = new LinkedList<>();
        for(int n=0; n<N; n++) shops.add(new Shop(sc.nextLong(), sc.nextInt()));
        Collections.sort(shops);

        long totalCost = 0;
        while(M>0){
            Shop bestShop = shops.poll();
            if(M <= bestShop.stock){
                totalCost += M * bestShop.price;
                M = 0;
            }else{
                totalCost += bestShop.stock * bestShop.price;
                M -= bestShop.stock;
            }
        }
        System.out.println(totalCost);





    }
}