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.Scanner;
import java.util.Arrays;

public class Main{

    public static void main(String[] args){
        Scanner scan = new Scanner(System.in);

        // input
        final int NUM = scan.nextInt();
        int goal = scan.nextInt();
        int[][] shop = new int[NUM][2];
        for(int i = 0; i < NUM; i++){
            shop[i][0] = scan.nextInt();
            shop[i][1] = scan.nextInt();
        }

        // sort
        Arrays.sort(shop, (a, b) -> Integer.compare(a[0], b[0]));

        // search
        long money = 0;
        for(int i = 0; i < NUM; i++){
            if(shop[i][1] <= goal){
                goal -= shop[i][1];
                money += (long)shop[i][0] * (long)shop[i][1];
                shop[i][1] = 0;
            } else {
                while(goal > 0 && shop[i][1] > 0){
                    goal -= 1;
                    money += shop[i][0];
                    shop[i][1] -= 1;
                }
            }
            if(goal <= 0){
                break;
            }
        }

        // answer
        System.out.println(money);

    }
}