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 java.math.*;

public class Main {

  public static void main(String[] args) {
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int m = sc.nextInt();
    // long[] a = new long[n];
    // long[] b = new long[n];

    // for (long i = 0; i < n; i++) {
    //   a[i] = sc.nextLong();
    //   b[i] = sc.nextLong();
    // }

    Map<Long,Long> map = new TreeMap<>();

    for (long i = 0; i < n; i++) {
      long a = sc.nextLong();
      long b = sc.nextLong();
      if (map.containsKey(a)) {
        map.put(a,map.get(a)+b);
      } else {
        map.put(a,b);
      }
    }

    long count = 0;
    long sum = 0;

    for (long i : map.keySet()) {
      if (count > m) {
        break;
      }

      if (count+map.get(i) <= m) {
        sum += i*map.get(i);
      } else {
        sum += (m-count)*i;
      }

      count += map.get(i);

    }

    System.out.println(sum);

  }
}