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.*;
public class Main{
	public static void main(String[] args){
    	Scanner sc = new Scanner(System.in);
      int n = sc.nextInt();
      int m = sc.nextInt();
      long sum = 0;
      int num = 0;
      Map<Integer,Integer> map = new HashMap<Integer,Integer>();
      ArrayList<Integer> list = new ArrayList<Integer>();
      for(int i = 0;i < n;i++){
        int a = sc.nextInt();
        int b = sc.nextInt();
        if(map.containsKey(a)){
          map.put(a,map.get(a)+b);
        }else{map.put(a,b);
        list.add(a);
        }
      }
      Collections.sort(list);
      for(int i = 0;i < list.size();i++){
      	int p = list.get(i);
        num += map.get(p);
        if(num < m)sum += (long)p*map.get(p);
        else {
          sum += (long)p*(map.get(p)-num+m);
          break;
        }
      }
      System.out.println(sum);
    }
}