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.List;
import java.util.Scanner;

public class Main {
  static class Data {
    public final long cost;
    public final long count;

    public Data(long cost, long count) {
      this.cost = cost;
      this.count = count;
    }
  }

  public static void main(String... args) {

    try (Scanner sc = new Scanner(System.in)) {
      long N = sc.nextLong();
      long M = sc.nextLong();
      List<Data> list = new ArrayList<>();
      for (int i = 0; i < N; ++i) {
        list.add(new Data(sc.nextLong(), sc.nextLong()));
      }
      list.sort(
          (a, b) -> {
            if (a.cost > b.cost) {
              return 1;
            }
            if (a.cost < b.cost) {
              return -1;
            }
            return 0;
          });

      long total = 0;
      for (Data d : list) {
        if (d.count < M) {
          total += d.cost * d.count;
          M -= d.count;
        } else {
          total += d.cost * M;
          break;
        }
      }
      System.out.println(total);
    }
  }

}