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.Comparator;
import java.util.Scanner;

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());
        int[][] ab = new int[n][2];
        for(int i=0;i<n;i++){
            ab[i][0] = Integer.parseInt(sc.next());
            ab[i][1] = Integer.parseInt(sc.next());
        }
        Arrays.sort(ab,new Comparator<int[]>(){

            @Override
            public int compare(int[] o1, int[] o2) {
                return o1[0]-o2[0];
            }
        }
        );
        int remain = m;
        long res = 0;
        int count = 0;
        for(int i=0;i<n;i++){
            if(remain==0)break;
            count = Math.min(remain,ab[i][1]);
            res+=(long)count*ab[i][0];
            remain-=count;
        }
        System.out.println(res);
        sc.close();
    }
    
}