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: Runtime Error)

import java.util.Scanner;
import java.util.List;
import java.util.ArrayList;
import java.util.Collections;
class Main{
  public static void main(String args[]){
    Scanner scan=new Scanner(System.in);
    int n=scan.nextInt();
    int m=scan.nextInt();
    List<Integer> a=new ArrayList<>();
    List<Integer> b=new ArrayList<>();
    List<Integer> cop=new ArrayList<>();
    for(int i=0;i<n;i++){
      a.add(scan.nextInt());
      cop.add(a.get(i));
      b.add(scan.nextInt());
    }
    Collections.sort(cop);
    long ans=0L;
    int i=-1;
    int sum=0;
    int index;
    int x=0;
    int y=0;
    while(sum<=m){
      i++;
      x=cop.get(i);
      index=a.indexOf(x);
      y=b.get(index);
      sum+=y;
      ans+=(long)(x*y);
    }
    ans-=(long)(x*(sum-m));
    System.out.println(ans);
  }
}