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.*;
 
public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        
        int N = Integer.parseInt(sc.next()); //N軒の店があり
        int M = Integer.parseInt(sc.next()); //M本の栄養ドリンクを買いたい
       
        long[] A = new long[N]; //iの店　1本Ai円の栄養ドリンクを
        int[] B = new int[N]; //Bi本買える
        //key：値段　value：本数
        Map<Long, Integer> map = new HashMap<>();
        
        for(int i=0; i<N; i++) {
            A[i] = Long.parseLong(sc.next());
            B[i] = Integer.parseInt(sc.next());
            if(map.containsKey(A[i])) {
                map.put(A[i], map.get(A[i])+B[i]);
            }else {
                map.put(A[i], B[i]);
            }
        }
        //昇順(安い順)に並べ替え
        Object[] key = map.keySet().toArray();
        Arrays.sort(key);
        
        int bin = M; //必要な本数
        long sum = 0;
        for(long k : map.keySet()) {

            if(bin > map.get(k)) {
                sum += k * map.get(k);
                bin -= map.get(k);
            }else {
                sum += k * bin;
                bin = 0;
            }
            if(bin <= 0) {
                break;
            }
        }
        System.out.println(sum);
        
    }
}