Problem:
Score : 300 points

Problem Statement

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.

Sample Input 1

2 5
4 9
2 4

Sample Output 1

12

With 12 yen, we can buy one drink at the first store and four drinks at the second store, for the total of five drinks. However, we cannot buy 5 drinks with 11 yen or less.

Sample Input 2

4 30
6 18
2 5
3 10
7 9

Sample Output 2

130

Sample Input 3

1 100000
1000000000 100000

Sample Output 3

100000000000000

The output may not fit into a 32-bit integer type.

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.io.*;
import java.util.*;

public class Main
{

	static OutputStream outputStream = System.out;
	static PrintWriter out = new PrintWriter(outputStream);

	static class node implements Comparable<node>
		{
			int data;int indx;
	
			node(int i,int o)
			{
				data=i;indx=o;
			}
			public int compareTo(node n)
			{
				if(this.data!=n.data) return this.data-n.data;
				return this.indx-n.indx;
			}
		}
	
	public static void main(String[] Rohan)throws IOException
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));

		String[] s=br.readLine().split(" ");
		int n=Integer.parseInt(s[0]),m=Integer.parseInt(s[1]);


		node[] a=new node[n];


		for(int i=0;i<n;i++)
		{
			s=br.readLine().split(" ");
			a[i]=new node(Integer.parseInt(s[0]),Integer.parseInt(s[1]));
		}

		Arrays.sort(a);


		long ans=0;

		for(int i=0;i<n;i++)
		{
			if(m<=0) break;

			long temp=a[i].data;
			temp*=Math.min(m,a[i].indx);
			m-=Math.min(m,a[i].indx);
			ans+=temp;
		}

		
		out.println(ans);
		out.close();
		

	}
}