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: Time Limit Exceeded)

using System;
using System.Linq;
class Program
{
    static void Main(string[] args)
    {
        string[] input = Console.ReadLine().Split(' ');
        int n = int.Parse(input[0]);
        int m = int.Parse(input[1]);

        //long[] d = new long[n];
        //input = Console.ReadLine().Split(' ');
        //for (int i = 0; i < n; i++)
        //{
        //    d[i] = long.Parse(input[i]);
        //}
        var items = Console.ReadLine().Split(' ').Select(long.Parse).ToList();
        items.Sort();
        items.Reverse();

        for (int i = 0; i < m; i++)
        {
            items[0] /= 2;
            if (n > 1)
            {
                if (items[0] < items[1])
                {
                    items.Sort();
                    items.Reverse();
                }
            }
        }

        long sum = 0;
        for (int i = 0; i < n; i++)
        {
            sum += items[i];
        }
        Console.WriteLine(sum);
    }
}