You are given a 0-indexed integer array `piles`, where `piles[i]` represents the number of stones in the `ith` pile, and an integer `k`. You should apply the following operation exactly `k` times:

- Choose any `piles[i]` and remove `floor(piles[i] / 2)` stones from it.

Notice that you can apply the operation on the same pile more than once.

Return the minimum possible total number of stones remaining after applying the `k` operations.

`floor(x)` is the greatest integer that is smaller than or equal to `x` (i.e., rounds `x` down).

Example 1:

    Input: piles = [5,4,9], k = 2
    Output: 12
    Explanation: Steps of a possible scenario are:
    - Apply the operation on pile 2. The resulting piles are [5,4,5].
    - Apply the operation on pile 0. The resulting piles are [3,4,5].
    The total number of stones in [3,4,5] is 12.

Example 2:

    Input: piles = [4,3,6,7], k = 3
    Output: 12
    Explanation: Steps of a possible scenario are:
    - Apply the operation on pile 2. The resulting piles are [4,3,3,7].
    - Apply the operation on pile 3. The resulting piles are [4,3,3,4].
    - Apply the operation on pile 0. The resulting piles are [2,3,3,4].
    The total number of stones in [2,3,3,4] is 12.

Constraints:

- `1 <= piles.length <= 105`
- `1 <= piles[i] <= 104`
- `1 <= k <= 105`