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.Numerics;
using System.Linq;
using System.Collections.Generic;
using System.Text;
using System.Collections;

namespace debug
{
    class main
    {
        static void Main(string[] args)
        {
            //問題クラスを展開
            ProgramC a = new ProgramC();
            a.main();//実行する

        }

    }

    class ProgramA
    {
        public void main()
        {
            //入力
            string[] s = Console.ReadLine().Split(' ');
            string[] t = Console.ReadLine().Split(' ');

            Console.WriteLine(int.Parse(s[0]) * int.Parse(s[1]) - int.Parse(s[0]) * int.Parse(t[1]) - int.Parse(s[1]) * int.Parse(t[0]) + int.Parse(t[1]) * int.Parse(t[0]));

        }
    }

    class ProgramB
    {
        public void main()
        {

            //入力
            string[] s = Console.ReadLine().Split(' ');
            int n = int.Parse(s[0]);
            int m = int.Parse(s[1]);
            int c = int.Parse(s[2]);

            int count = 0;

            string[] b = Console.ReadLine().Split(' ');
            for (int i = 0; i < n; i++)
            {
                string[] a = Console.ReadLine().Split(' ');
                int sum = c;
                for(int j = 0;j < m;j++)
                    sum += int.Parse(a[j]) * int.Parse(b[j]);

                if (sum > 0)
                    count++;
            }

            

            //答え出力
            Console.WriteLine(count);
            

        }
    }

    class ProgramC
    {
        public void main()
        {

            //入力
            string[] ss = Console.ReadLine().Split(' ');
            int n = int.Parse(ss[0]);
            long m = long.Parse(ss[1]);

            //i文字目までにACが何個あるか記録する
            int[] AC = new int[n + 1];
            int count = 0;

            List<Task> list = new List<Task>();

            for(int i = 0; i < n;i++)
            {
                string[] s = Console.ReadLine().Split(' ');
                list.Add(new Task(long.Parse(s[0]), long.Parse(s[1])));
            }

            var sort = list.OrderBy(x => x.a);
            long ans = 0;

            foreach(var s in sort)
            {
                if(m > s.b)
                {
                    ans += s.a * s.b;
                    m -= s.b;
                }
                else
                {
                    ans += s.a * m;
                    break;
                }

            }

            Console.WriteLine(ans);

        }


    }

    class Task
    {
        public long a;
        public long b;
        public Task(long a,long b)
        {
            this.a = a;
            this.b = b;
        }

    }




    class ProgramD
    {
        public void main()
        {

            //入力
            int n = int.Parse(Console.ReadLine());
            long MOD = 1000000000 + 7;
            long[,,,] dp = new long[n, 4, 4, 4];

            

            //初項
            for (int k = 0; k < 4; k++)//2つ前
                for (int l = 0; l < 4; l++)//一つ前
                    for (int m = 0; m < 4; m++)//今
                    {
                        if ((k == 0 && l == 2 && m == 1) || (k == 0 && l == 1 && m == 2) || (k == 2 && l == 0 && m == 1))
                            continue;

                        dp[2, k, l, m] = 1;
                    }


            //動的計画法
            for (int i = 3; i < n; i++)//4文字目から見る
                for (int j = 0; j < 4; j++)//3つ前
                    for (int k = 0; k < 4; k++)//2つ前
                        for (int l = 0; l < 4; l++)//一つ前
                            for (int m = 0; m < 4; m++)//今参照してる列
                            {
                                // [A, C, G, T]
                                // _AG -> C (0,2,1)
                                // _AC -> G (0,1,2)
                                // _GA -> C (2,0,1)
                                // AG_ -> C (0,2,X,1)
                                // A_G -> C (0,X,2,1)
                                if ((k == 0 && l == 2 && m == 1) || (k == 0 && l == 1 && m == 2) ||
                                    (k == 2 && l == 0 && m == 1) || (j == 0 && k == 2 && m == 1) || (j == 0 && l == 2 && m == 1))
                                    continue;

                                dp[i, k, l, m] += dp[i - 1, j, k, l];
                                dp[i, k, l, m] %= MOD;


                            }

            long ans = 0;

            for (int k = 0; k < 4; k++)//2つ前
                for (int l = 0; l < 4; l++)//一つ前
                    for (int m = 0; m < 4; m++)//今
                    {
                        ans += dp[n -1, k, l, m];
                        ans %= MOD;
                    }

            //答え出力
            Console.WriteLine(ans);

        }

    }


}