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 static System.Console;

namespace c {

static class Extensions {
   public static long ToInt(this string str) {
      return Int64.Parse(str);
   }
}

class Program {
   public static void Main(string[] args) {
      var tarr = ReadLine().Split(' ').Select(x => x.ToInt()).ToArray();
      var n = tarr[0];
      var m = tarr[1];
      
      var arr = new long[n][];
      for (int i = 0; i < n; i++) {
         arr[i] = ReadLine().Split(' ').Select(x => x.ToInt()).ToArray();
      }

      arr = arr.OrderBy(x => x[1]).ToArray();
      
      long count = 0;
      long okane = 0;
      for (int i = 0; i < n; i++) {
         for (int j = 0; j < arr[i][1]; j++) {
            if (count != m) {
               count++;
               okane += arr[i][0];
            } else {
               goto label;
            }
         }

         // if (arr[i][1] + count < m) {
         //    count += arr[i][1];
         //    okane += arr[i][0] * arr[i][1];
         // } else {
         //    for (int j = 0; j < arr[i][1]; j++) {
         //       if (count != m) {
         //          count++;
         //          okane += arr[i][0];
         //       } else {
         //          goto label;
         //       }
         //    }
         // }
      }
      label:

      WriteLine(okane);
   }
}

}