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.Comparator;
import java.util.List;
import java.util.Scanner;
 
public class Main {
 
  public static void main(final String[] ars) {
    final Scanner scanner = new Scanner(System.in);
    final int N = Integer.parseInt(scanner.next());
    final int M = Integer.parseInt(scanner.next());
    final List<Shop> list = new ArrayList<>();
    for (int i = 0; i < N; i++) {
      list.add(new Shop(Long.parseLong(scanner.next()), Long.parseLong(scanner.next())));
    }
    list.sort(Comparator.comparingLong(e -> e.price));
 
    long price = 0;
    long count = 0;
    for (final Shop shop : list) {
      if (count + shop.count >= M) {
        price += (M - count) * shop.price;
        break;
      }
      count += shop.count;
      price += shop.count * shop.price;
    }
    System.out.println(price);
  }
}
 
 
class Shop {
  Long price;
  Long count;
 
  public Shop(final Long price, final Long count) {
    this.price = price;
    this.count = count;
  }
}