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 asc(const void *a, const void *b) {
  return *(int *)a - *(int *)b;
}


int main(){
    int n,m;
    scanf("%d %d",&n,&m);
    int val[n];
    long ans;
    for(int i=0;i<n;i++){
        scanf("%d ",&val[i]);
    }
    qsort(val,n,sizeof(int),asc);
    for(int i=0;i<m;i++){
        val[n-1]=val[n-1]/2;
        int j=n-1;
        while(j>0&&val[j-1]>val[j]){
            int tmp=val[j];
            val[j]=val[j-1];
            val[j-1]=tmp;
            j--;
        }
    }
    ans=0;
    for(int i=0;i<n;i++){
        ans=ans+(long)val[i];
    }
    printf("%ld",ans);
    return 0;
}