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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {
    public static void debug(String str) {
        System.out.println(str);
    }

    public static void main(String[] args) {

        int N;
        int M;
        long[] Ai;

        Comparator<Double> comp = Comparator.comparingDouble(d->d.doubleValue());
        PriorityQueue<Double> pq = new PriorityQueue<>(comp.reversed());

        try (Scanner sc = new Scanner(System.in)) {
            N = sc.nextInt();
            M = sc.nextInt();

            Ai = new long[N];
            for (int i = 0; i < Ai.length; i++) {
                Ai[i] = sc.nextLong();
            }
        }
        for(long l : Ai){
            pq.add((double) l);
        }
        Double d ;
        while(M >0){
            d = pq.poll();
            double temp = d/2;
            pq.add(temp);
            M--;
        }
        long result = 0;
        while((d = pq.poll()) != null){
            result += Math.floor(d);
        }
        System.out.println(result);
    }
}