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.Scanner;
import java.util.Arrays;
import java.util.Comparator;
public class Main{
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		int n=sc.nextInt();
		int m=sc.nextInt();
		Point[] p=new Point[n];
		int count=0;
		long ans=0;
		for(int i = 0; i<n; i++){
			long a=sc.nextLong();
			int b = sc.nextInt();
			p[i]=new Point(a,b);
		}
		Arrays.sort(p,Comparator.comparing(Point::getX));
		while(m>0){
			if(p[count].y<m){
				m-=p[count].y;
				ans+=p[count].x*p[count].y;
			}else{
				ans+=p[count].x*m;
				m=0;
			}
			count++;
		}
		System.out.println(ans);
	}
	static class Point{
		long x;
		int y;
		public Point(long x, int y){
			this.x=x;
			this.y=y;
		}
		
		public long getX(){
			return x;
		}
		 
		public int getY(){
			return y;
		}
	}
}