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: Time Limit Exceeded)

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

int main(void){
    int N,M;
    unsigned long int A[100001],sum=0;
    scanf("%d%d",&N,&M);
    int i,j,tmp;
    
    for(i=0;i<N;i++) scanf("%ld",&A[i]);
    
    
    for(i=0;i<N-1;i++){
        for(j=0;j<N-1;j++){
            if(A[j]<A[j+1]){
                tmp=A[j];
                A[j]=A[j+1];
                A[j+1]=tmp;
            }
        }
    }
    
    int a,b;
    for(a=0;a<M;a++){
        if(A[0]/2>A[1]&&a+1<M){
            A[0]/=4;
            a++;
        }else if(A[0]/2>A[1]&&a+1<M){
            A[0]/=2;
            A[1]/=2;
            a++;
        }else{
            A[0]/=2;
            for(i=0;i<N-1;i++){
                for(j=0;j<N-1;j++){
                    if(A[j]<A[j+1]){
                        tmp=A[j];
                        A[j]=A[j+1];
                        A[j+1]=tmp;
                    }
                }
            }
        }
    }
    for(i=0;i<N;i++)sum+=A[i];
    printf("%ld\n",sum);
    return 0;
}