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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Linq;

namespace C___Energy_Drink_Collector
{
    class Program
    {
        static void Main(string[] args)
        {
            var NM = Console.ReadLine().Split().Select(int.Parse).ToArray();
            int N = NM[0], M = NM[1];
            var AiBi = Enumerable.Range(1, N).Select(_ => Console.ReadLine().Split().Select(int.Parse).ToArray()).OrderBy(x => x[0]).ToArray();

            long count = 0, sum = 0;
            for (int i = 0; i < N; i++)
            {
                long A = AiBi[i][0], B = AiBi[i][1];

                if (count + B <= M)
                {
                    sum += A * B;
                    count += B;
                }
                else
                {
                    sum += A * (M - count);
                    break;
                }
            }

            Console.WriteLine(sum);
        }
    }
}