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.io.InputStream;
import java.io.PrintStream;
import java.util.*;

public class Main {

  public static void main(String[] args) {
    solve(System.in, System.out);
  }

  private static class Pair {
    long price;
    long num;
  }

  static void solve(InputStream is, PrintStream os) {
    Scanner sc = new Scanner(is);

    /* read */
    int n = sc.nextInt();
    int m = sc.nextInt();

    Queue<Pair> pairs = new PriorityQueue<>(Comparator.comparing(p -> p.price));

    for (int i = 0; i < n; i++){
      Pair pair = new Pair();
      pair.price = sc.nextLong();
      pair.num = sc.nextLong();
      pairs.add(pair);
    }

    int remain = m;
    long cost = 0;
    while (!pairs.isEmpty()) {
      Pair reasonable = pairs.remove();
      if (remain > reasonable.num ) {
        cost += reasonable.price * reasonable.num;
        remain -= reasonable.num;
      } else {
        cost += reasonable.price * remain;
        os.println(cost);
        return;
      }
    }
  }

}