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: Accepted)

/*
* @Author Silviase(@silviasetitech)
* For ProCon
*/

import java.util.*;
import java.lang.*;
import java.math.*;

class Main{
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int[] a = new int[n];
        int[] b = new int[n];
        for (int i = 0; i < n; i++) {
            a[i] = sc.nextInt();
            b[i] = sc.nextInt();
        }

        abc121cEnergyDrinkCollectorSolve(n,m,a,b);
        sc.close();
    }

    public static void abc121cEnergyDrinkCollectorSolve(int n, int m, int[] a, int[] b) {
        long sum = 0;
        int numcount = 0;
        int ite = 0;
        boolean bool = true;
        Integer[] id = new Integer[n];
        for (int i = 0; i < n; i++) {
            id[i] = i;
        }
        Arrays.setAll(id, i -> i);
        Arrays.sort(id, Comparator.comparingInt(i -> a[i]));
        while(bool){
            if (numcount + b[id[ite]] < m) {
                sum += (long)b[id[ite]] * (long)a[id[ite]];
                numcount += b[id[ite]];
            } else {
                sum += (long)(m-numcount) * (long)a[id[ite]];
                bool = false;
            }
            ite++;
        }
        System.out.println(sum);
    }
}