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.*;

public class Main {

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

        int N = Integer.parseInt(sc.next()); //N軒の店があり
        int M = Integer.parseInt(sc.next()); //M本の栄養ドリンクを買いたい

        long[] A = new long[N]; //iの店　1本Ai円の栄養ドリンクを
        int[] B = new int[N]; //Bi本買える
        //key：値段　value：本数
        Map<Long, Integer> map = new TreeMap<>();

        for(int i=0; i<N; i++) {
            A[i] = Long.parseLong(sc.next());
            B[i] = Integer.parseInt(sc.next());
            if(map.containsKey(A[i])) {
                map.put(A[i], map.get(A[i])+B[i]);
            }else {
                map.put(A[i], B[i]);
            }
        }

        Long count = 0l;
        Long total = 0l;
        for (Long str : map.keySet()) {
            for (int i = 0; i < map.get(str); i++) {
                if(count >= M) break;
                total += str;
                count++;
            }
        }
        System.out.println(total);
    }
}