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 (Clang 3.8.0) (judge verdict: Accepted)

#include<cstdio>
#include<stdio.h>
#include<math.h>
#include<queue>
#include<iostream>
#include<string>
using namespace std;
int main()
{
	priority_queue<long long>q;
	int m,n,op;
	long long sum=0;
	cin>>n>>m;
	for(int i=0;i<n;i++)
    {
		cin>>op;
		q.push(op);
	}
	while(m--)
	{
		op=q.top()/2;
		q.pop();
		q.push(op);
	}
	while(!q.empty())
	{
		sum+=q.top();
		q.pop();
	}
	cout<<sum<<endl;
	return 0;
}