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: Time Limit Exceeded)

import java.util.*;

class Main{
	public static void main(String args[]){
		Scanner sc = new Scanner(System.in);
		long n = sc.nextLong();
		long m = sc.nextLong();
		ArrayList<Long> a = new ArrayList<Long>();
		ArrayList<Long> b = new ArrayList<Long>();
		ArrayList<Long> a2 = new ArrayList<Long>();
		for(int i = 0; i < n; i++){
			a.add(sc.nextLong());
			a2.add(a.get(i));
			b.add(sc.nextLong());
		}
		Collections.sort(a2);
		long ans = 0;
		long ansb = 0;
		long i = a2.get(0);
		while(ansb != m){
			int idx = a.indexOf(i);
			if(idx == -1){
				a2.remove(0);
				i = a2.get(0);
			}else{
				long geta = a.get(idx);
				long getb = b.get(idx);
				if(ansb+getb >= m){
					ans =(long)ans+geta*(m-ansb);
					ansb = m;
				}else{
					ansb+=getb;
					ans =(long)ans+geta*getb;
					a.remove(idx);
					b.remove(idx);
					a2.remove(0);
					i = a2.get(0);
				}
			}
		}
		System.out.print(ans);
	}
}