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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

typedef struct priceHeap
{
    long int *product;
    int head;
    int tail;
} heap;

void initialization(heap *h, int inputN)
{
    h->product = (long int *)malloc(sizeof(long int) * inputN);
    for (int index = 0; index < inputN; index++)
    {
        h->product[index] = 0;
    }
    h->head = 0;
    h->tail = 0;
}

int putInHeap(heap *h, long int price, int inputN)
{
    if (h->tail >= inputN)
    {
        return -1;
    }
    else
    {
        h->product[h->tail] = price;
        h->tail++;
        return 0;
    }
}

void pushDown(heap *h, int firstIndex)
{
    int parentIndex = firstIndex, childIndex;
    long int parent, child;

    while (parentIndex * 2 < h->tail - 1)
    {
        if (parentIndex == 0)
        {
            childIndex = 1;
        }
        else
        {
            childIndex = parentIndex * 2;
        }
        if (h->product[childIndex] < h->product[childIndex + 1])
        {
            childIndex++;
        }
        parent = h->product[parentIndex];
        child = h->product[childIndex];
        if (parent < child)
        {
            h->product[parentIndex] = child;
            h->product[childIndex] = parent;
            parentIndex = childIndex;
        }
        else
        {
            break;
        }
    }
}

void heapSort(heap *h)
{
    for (int count = (h->tail - 1) / 2; count >= 0; count--)
    {
        pushDown(h, count);
    }
}

long long int totalPrice(heap *h)
{
    long long int total = 0;
    for (int index = 0; index < h->tail; index++)
    {
        total += h->product[index];
    }
    return total;
}

int main(void)
{
    int inputN, inputM, flag = 0;
    long int *inputA;
    long long int total = 0;
    heap list;
    scanf("%d%d", &inputN, &inputM);
    inputA = (long int *)malloc(sizeof(long int) * inputN);
    initialization(&list, inputN);
    for (int count = 0; count < inputN; count++)
    {
        scanf("%ld", &inputA[count]);
        putInHeap(&list, inputA[count], inputN);
    }
    for (int count = 0; count < inputM; count++)
    {
        heapSort(&list);
        list.product[0] /= 2;
    }
    total = totalPrice(&list);
    printf("%lld", total);

    free(inputA);
    free(list.product);

    return 0;
}