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: Wrong Answer)

using System;
using System.Linq;
using System.Collections.Generic;

class Cmondai
{
    static void Main()
    {
        var nm = Console.ReadLine().Split(' ').Select(x => Int32.Parse(x)).ToArray();
        var list = new List<long[]>();
        list.Add(new long[3] { 0, long.MaxValue, 0 });
        for (int i = 0; i < nm[0]; ++i)
        {
            var ab = Console.ReadLine().Split(' ').Select(x => Int64.Parse(x)).ToArray();
            list.Add(new long[3] { i + 1, ab[0], ab[1] });
        }
        var ans = long.MaxValue;
        for (int i = 0; i < (Math.Pow(2, nm[0])); i++)
        {
            var mondaiCount = 0L;
            for (int j = 0; j < nm[0]; j++)
            {
                mondaiCount++;
                if ((1 & i >> j) == 1)
                {
                    mondaiCount += list[j][2] - 1;
                }
            }
            //買うぶんを超えてしまう
            if (nm[1] < mondaiCount)
            {
                continue;
            }
            else
            {
                var flg = false;
                var total = 0L;
                var last = 0;
                //足りないぶんを連番で補充する
                for (int j = nm[0] - 1; 0 <= j; --j)
                {
                    //全て
                    if ((1 & i >> j) == 1)
                    {
                        flg = true;
                        last = j;
                        total += list[j][1] * list[j][2];
                    }
                    else if (flg)
                    {
                        total += list[j][1];
                    }
                }
                var tarinai = nm[1] - mondaiCount;
                for (int j = last + 1; j < nm[0]; ++j)
                {
                    if (list[j][2] <= tarinai)
                    {
                        total += list[j][1] * list[j][2];
                        tarinai -= list[j][2];
                    }
                    else
                    {
                        total += list[j][1] * tarinai;
                        tarinai = 0;
                    }
                    if (tarinai == 0)
                    {
                        ans = Math.Min(ans, total);
                        break;
                    }
                }
            }
        }
        Console.WriteLine(ans);
    }
}