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.Arrays;
import java.util.Comparator;
import java.util.Scanner;
class pair{
	int cost;
	int val;
	
	pair(int a,int b)
	{
		cost=a;
		val=b;
	}
}
class compare implements Comparator<pair>{

	@Override
	public int compare(pair o1, pair o2) {
		if(o1.cost>o2.cost)
		{
			return 1;
			
		}
		else if(o1.cost<o2.cost){
			return -1;
			
		}
		else {
			return 0;
		}
	}
	
}
public class Main {

	public static void main(String args[])
	{
		Scanner scan=new Scanner(System.in);
		int n=scan.nextInt();
		long m=scan.nextInt();
		pair[] x=new pair[n];
		for(int i=0;i<n;i++)
		{
			x[i]=new pair(scan.nextInt(),scan.nextInt());
		}
		Arrays.sort(x,new compare());
		long cur=0;
		long total=0;
		int index=0;
		while(cur<m)
		{
			total+=(x[index].cost*Math.min(m-cur,x[index].val));
			cur+=x[index].val;
			index++;
		}
		System.out.println(total);
	}
}