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>

void swap(long *x,long *y){
    long z;
    z=*x;
    *x=*y;
    *y=z;
}

int max(long a,long b){
    if(a>b)return a;
    else return b;
}

void pushdown(long* a,int n){
    int i=1;
    while(1){
        if(a[i]>=max(a[2*i],a[2*i+1]))break;
        if(a[2*i]>a[2*i+1]){
            swap(&a[i],&a[2*i]);
            i*=2;
        }else{
            swap(&a[i],&a[2*i+1]);
            i=2*i+1;
        }
        if(i*2>=n)break;
    }
}

void add_num(long*a,int i){
    while(i>1){
        if(a[i]<=a[i/2])break;
        swap(&a[i],&a[i/2]);
        i/=2;
    }
}

int main(void){
    int N,M,tmp;
    scanf("%d%d",&N,&M);
    long a[100000];
    for(int i=0;i<100000;i++)a[i]=0;
    for(int i=1;i<N+1;i++){
        scanf("%ld",&a[i]);
        add_num(a,i);
    }
    for(int i=0;i<M;i++){
        a[1]=a[1]/2;
        pushdown(a,N);
    }
    long count=0;
    for(int i=1;i<N+1;i++){
        count+=a[i];
    }
    printf("%ld\n",count);
    return 0;
}