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.*;
import static java.lang.System.*;

class Drink implements Comparable<Drink>{
    long p;
    long n;
    
    Drink(long _p, long _n) {
        p = _p;
        n = _n;
    }
    
    public int compareTo(Drink o) {
        if(p > o.p) return 1;
        else if (p < o.p) return -1;
        else return 0;
    }
}

public class Main {
    public static void main(String[] args) throws Exception {
        Scanner sc = new Scanner(in);
        
        int n = sc.nextInt();
        long m = sc.nextLong();
        
        Drink[] d = new Drink[n];
        for(int i=0; i<n; i++) 
            d[i] = new Drink(sc.nextLong(), sc.nextLong());
        
        Arrays.sort(d);
        
        long price = 0;
        long num = 0;
        for(int i=0; i<n; i++) {
            if(num >= m) break;
            else if(num + d[i].n >= m) {
                price += (m - num) * d[i].p;
                break;
            }
            else {
                price += d[i].n * d[i].p;
                num += d[i].n;
            }
        }
        
        out.println(price);
    }
}