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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <string>
#include <algorithm>
#include <utility>
using namespace std;

int main(){
    int n,m;
    cin >> n >> m;
    int a[n];

    for(int i = 0;i<n;i++){
        cin >> a[i];
    }
    sort(a,a+n);
    int lowest = 0;
    int highest = n-1;
    for(int i = 0;i < m;i++){
        a[n-1] /= 2;
        while(true){
            int b = (highest + lowest)/2;
            if(highest - lowest == 0){
                swap(a[n-1],a[lowest]);
              	//for(int j = 0;j<n;j++)cout << a[j] << endl;
                break;
              	
            }
            if(a[n-1] > a[b]){
                lowest = b+1;
            }else{
                highest = b;
            }
          //cout << a[b] << endl;
        }
    }
    unsigned long long sum = 0;
    for(int i = 0;i < n;i++){
        sum += a[i];
    }
    cout << sum << endl;
}