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.*;
public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);
		
		int N = sc.nextInt();
		
		int M = sc.nextInt();
		
		long[] A = new long[N];
		
		long[] B = new long[N];
		
		for(int i=0;i<N;i++){
			
			A[i] = sc.nextLong();
			
			B[i] = sc.nextLong();
		}
		
		for(int i=0;i<N-1;i++){
			
			for(int j=i+1;j<N;j++){
				
				if(A[j]<A[i]){
					
					long tempA = A[i];
					
					A[i] = A[j];
					
					A[j] = tempA;
					
					long tempB = B[i];
					
					B[i] = B[j];
					
					B[j] = tempB;
									
				}
			}
		}
		
		long ans = 0;
		
		long k = 0;
		
		int i = 0;
		
		while(k<M){
			
			if(k+B[i]<=M){
				
				ans += A[i]*B[i];
				
				k += B[i];
								
			}else{
				
				for(;k<M;k++){
					
					ans += A[i];
									}
			}
			i++;
		}
		System.out.println(ans);
				
	}
	
	
}