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.io.*;
import java.util.*;

@SuppressWarnings("unused")
class Main {

    // Generated by 1.1.5 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
    public static void main(String[] args) throws Exception {
        final Scanner sc = new Scanner(System.in);
        long N;
        N = sc.nextLong();
        long M;
        M = sc.nextLong();

        PriorityQueue<Integer> A = new PriorityQueue<>(Comparator.reverseOrder());
        for(int i = 0 ; i < N ; i++){
            A.add(sc.nextInt());
        }
        solve(N, M, A);
    }

    static void solve(long N, long M, PriorityQueue<Integer> A){
        for (int i = 0; i < M; i++) {
            int max = A.poll();
            A.offer(max/2);
        }
        long ans = 0;
        for (int i : A) {
            ans += i;
        }
        System.out.println(ans);
    }
}