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.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Scanner;
import static java.util.Comparator.*;


public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = Integer.parseInt(sc.next());
        int M = Integer.parseInt(sc.next());
        ArrayList<ArrayList<Long>> a = new ArrayList<>();
        for(int i = 0; i < N ;i++){
            ArrayList<Long> a2 = new ArrayList<>();
            a2.add(Long.parseLong(sc.next()));
            a2.add(Long.parseLong(sc.next()));
            a.add(a2);
        }
        a.sort(comparing(x->x.get(0)));
        long count = 0l;
        for(ArrayList<Long> ae : a){
            while(M > 0 && ae.get(1) > 0){
                count += ae.get(0);
                ae.set(1, ae.get(1)-1);
                M--;
            }
        }
        System.out.println(count);
    }

}