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: Wrong Answer)

import java.util.Arrays;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;

public class Main {

	public static void main(String args[]) {
		Scanner sc = new Scanner(System.in);
		long n=sc.nextLong();
		long m= sc.nextLong();
		long count=0;
		long price=0;
		Map<Long,Long> map = new HashMap<>();
		for(int i=0;i<n;i++) {
				long key= sc.nextLong();
				long value = sc.nextLong();
				if(map.containsKey(key)) {
			map.put(key,value+map.get(key));
				}else {
					map.put(key,value);
				}
		}
		Object[] mapkey = map.keySet().toArray();
		Arrays.sort(mapkey);
		for( long nkey: map.keySet()) {
			long tempCount=map.get(nkey);
			m-=tempCount;
			if(m>0) {
				count+=tempCount*nkey;
			}else{
				count+=nkey+(m+tempCount);
				System.out.println(count);
				break;
			}


		}
    }

}