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 Go (1.6) (judge verdict: Wrong Answer)

package main

import (
	"bufio"
	"fmt"
	"os"
	"sort"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}

func nextFloat() float64 {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return float64(i)
}

func main() {
	sc.Split(bufio.ScanWords)
	n := nextInt()
	m := nextInt()
	prices := make([]float64, n)
	for i := 0; i < n; i++ {
		prices[i] = nextFloat()
	}
	sort.Float64s(prices)
	for i := 0; i < m; i++ {
		half := prices[n-1] / 2.0
		for i := 2; i <= n; i++ {
			if i == 2 && half > prices[n-2] {
				prices[n-1] = half
				break
			} else if half > prices[n-i] {
				prices = append(prices[:n-i+1], append([]float64{half}, prices[n-i+1:n-1]...)...)
				break
			}
		}
	}
	var sum int
	for i := 0; i < n; i++ {
		sum += int(prices[i])
	}
	fmt.Println(sum)
}