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 JavaScript (Node.js 12.16.1) (judge verdict: Time Limit Exceeded)

"use strict"
const create = () => {
    const input = require("fs").readFileSync("/dev/stdin", "utf8").trim().split("\n");
    const res = {
        "list": input, "index": 0, "max": input.length,
        "hasNext": function() {return this.index < this.max},
        "next": function() { if (!this.hasNext()) {throw "ArrayIndexOutOfBoundsException";}else{return this.list[this.index++];}}
    };
    return res;
}
const o = create();
const next = () => o.next()
const nextInt = () => parseInt(o.next())
const nextStrArray = () => o.next().split(" ")
const nextIntArray = () => o.next().split(" ").map(el => parseInt(el))
const nextCharArray = () => o.next().split("")
const hasNext = () => o.hasNext()
const myout = (...x) => console.log(...x)

class MaxHeap {
    arr
    constructor() {
        this.arr = []
    }
    build() {
        for (let i = this.arr.length - 1; i >= 0; i--) {
            this.heapify(i, this.arr.length)
        }
    }
    heapify(i, len) {
        let l = 2 * i + 1
        let r = 2 * i + 2
        let largest = i

        // Condition 'l <= len' & 'r <= len' is for the case of leaf nodes.
        // On leaf nodes, swap() should NOT happen so this condition skips these cases.
        if (l <= len && this.arr[l] > this.arr[i]) {
            largest = l
        }
        if (r <= len && this.arr[r] > this.arr[i]) {
            largest = r
        }
        if (largest !== i) {
            this.swap(i, largest)
            this.heapify(largest, len)
        }
    }
    swap(i, j) {
        [this.arr[i], this.arr[j]] = [this.arr[j], this.arr[i]]
    }
    push(data) {
        this.arr.push(data)
        this.build()
    }
    pop() {
        let elm = this.arr.shift()
        this.build()
        return elm
    }
}

const Main = () => {
    let [N, M] = nextIntArray()
    const a = nextIntArray()
    a.sort((a,b) => b-a)
    let mh = new MaxHeap()
    for(let el of a){
        mh.push(el)
    }
    while(M>0){
        let tmp = mh.pop()
        tmp = Math.floor(tmp/2)
        mh.push(tmp)
        M--
    }
    let sum = 0
    for(let el of mh.arr){
        sum += el
    }
    console.log(sum)
}
Main()