Problem:
Score : 400 points

Problem Statement

Takahashi is going to buy N items one by one.

The price of the i-th item he buys is A_i yen (the currency of Japan).

He has M discount tickets, and he can use any number of them when buying an item.

If Y tickets are used when buying an item priced X yen, he can get the item for \frac{X}{2^Y} (rounded down to the nearest integer) yen.

What is the minimum amount of money required to buy all the items?

Constraints

All values in input are integers.

1 \leq N, M \leq 10^5

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N M
A_1 A_2 ... A_N

Output

Print the minimum amount of money required to buy all the items.

Sample Input 1

3 3
2 13 8

Sample Output 1

9

We can buy all the items for 9 yen, as follows:

Buy the 1-st item for 2 yen without tickets.

Buy the 2-nd item for 3 yen with 2 tickets.

Buy the 3-rd item for 4 yen with 1 ticket.

Sample Input 2

4 4
1 9 3 5

Sample Output 2

6

Sample Input 3

1 100000
1000000000

Sample Output 3

0

We can buy the item priced 1000000000 yen for 0 yen with 100000 tickets.

Sample Input 4

10 1
1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000 1000000000

Sample Output 4

9500000000

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

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

namespace D
{
  class Program
  {
    static void Main(string[] args)
    {
      Console.WriteLine(Run(args));
    }

    private static long Run(string[] args)
    {
      var input = Console.ReadLine().Split(' ').Select(int.Parse).ToArray();
      var N = input[0];
      var M = input[1];

      var items = Console.ReadLine().Split(' ').Select(double.Parse).ToList();
      items.Sort();
      items.Reverse();

      var ret = 0L;
      while (M > 0)
      {
        items[0] /= 2;
        M--;
        if (M == 0) break;

        for (int i = 1; i < N; i++)
        {
          if (items[0] > items[i])
          {
            items.Sort();
            items.Reverse();
            break;
          }
          items[i] /= 2;
          M--;
          if (M == 0) break;
        }
      }

      ret = items.Select(d => (long)d).Sum();

      return ret;
    }
  }
}