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 Java (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.math.BigInteger;
import java.util.*;
import java.util.stream.Collectors;

public class Main {
  public static void main(String[] args){
    Scanner sc = new Scanner(System.in);
    int n = sc.nextInt();
    int m = sc.nextInt();
    List<Long> a = new ArrayList<>();
    for (int i = 0; i < n; i++) {
      a.add(sc.nextLong());
    }
    if (n == 1) {
      long p = a.get(0);
      for (int i = 0; i < m; i++) {
        p = p / 2;
      }
      System.out.println(p);
      return;
    }
    a.sort(Comparator.reverseOrder());
    for (int i = 0; i < m; i++) {
      Long aLong = a.get(0);
      long newLong = aLong / 2;
      a.set(0, newLong);
      if (newLong < a.get(1)) {
        a.sort(Comparator.reverseOrder());
      }
    }
    System.out.println(a.stream().reduce(0L, Long::sum));
  }
}