<p>You are given an integer array <code>nums</code>, an integer <code>k</code>, and an integer <code>multiplier</code>.</p>

<p>You need to perform <code>k</code> operations on <code>nums</code>. In each operation:</p>

<ul>
	<li>Find the <strong>minimum</strong> value <code>x</code> in <code>nums</code>. If there are multiple occurrences of the minimum value, select the one that appears <strong>first</strong>.</li>
	<li>Replace the selected minimum value <code>x</code> with <code>x * multiplier</code>.</li>
</ul>

<p>After the <code>k</code> operations, apply <strong>modulo</strong> <code>10<sup>9</sup> + 7</code> to every value in <code>nums</code>.</p>

<p>Return an integer array denoting the <em>final state</em> of <code>nums</code> after performing all <code>k</code> operations and then applying the modulo.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [2,1,3,5,6], k = 5, multiplier = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">[8,4,6,5,6]</span></p>

<p><strong>Explanation:</strong></p>

<table>
	<tbody>
		<tr>
			<th>Operation</th>
			<th>Result</th>
		</tr>
		<tr>
			<td>After operation 1</td>
			<td>[2, 2, 3, 5, 6]</td>
		</tr>
		<tr>
			<td>After operation 2</td>
			<td>[4, 2, 3, 5, 6]</td>
		</tr>
		<tr>
			<td>After operation 3</td>
			<td>[4, 4, 3, 5, 6]</td>
		</tr>
		<tr>
			<td>After operation 4</td>
			<td>[4, 4, 6, 5, 6]</td>
		</tr>
		<tr>
			<td>After operation 5</td>
			<td>[8, 4, 6, 5, 6]</td>
		</tr>
		<tr>
			<td>After applying modulo</td>
			<td>[8, 4, 6, 5, 6]</td>
		</tr>
	</tbody>
</table>
</div>

<p><strong class="example">Example 2:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [100000,2000], k = 2, multiplier = 1000000</span></p>

<p><strong>Output:</strong> <span class="example-io">[999999307,999999993]</span></p>

<p><strong>Explanation:</strong></p>

<table>
	<tbody>
		<tr>
			<th>Operation</th>
			<th>Result</th>
		</tr>
		<tr>
			<td>After operation 1</td>
			<td>[100000, 2000000000]</td>
		</tr>
		<tr>
			<td>After operation 2</td>
			<td>[100000000000, 2000000000]</td>
		</tr>
		<tr>
			<td>After applying modulo</td>
			<td>[999999307, 999999993]</td>
		</tr>
	</tbody>
</table>
</div>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= k &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= multiplier &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution:
    def getFinalState(self, nums: List[int], k: int, multiplier: int) -> List[int]:
        if multiplier == 1:
            return nums

        n = len(nums)
        MODULO = int(1e9) + 7
        heap = sorted((num, i) for i, num in enumerate(nums))[:k]
        maxNum = heap[-1][0]
        heapify(heap)

        for j in range(k):
            num, i = heappop(heap)
            nums[i] = num * multiplier
            heappush(heap, (nums[i], i))
            if num > maxNum:
                break
        
        remaining = k - j - 1
        if remaining == 0:
            return [num % MODULO for num in nums]


        div, mod = divmod(remaining, n)
        smallest = nsmallest(mod, heap)
        indices = set(i for num, i in smallest)
        potencies = [div + (1 if i in indices else 0) for i in range(n)]
        
        multiplierPotencies = {}
        @cache
        def getMultiPotency(multiplier, potency):
            if potency == 0:
                return 1

            if multiplier not in multiplierPotencies:
                multiplierPotencies[multiplier] = [(1, 0), (multiplier, 1)]
                multipliers = multiplierPotencies[multiplier]
                while multipliers[-1][1] < potency:
                    multi, pot = multipliers[-1]
                    multipliers.append((multi * multi % MODULO, pot * 2))
            else:
                multipliers = multiplierPotencies[multiplier]
            multi = 1
            while potency > 1:
                l, r = 0, len(multipliers)
                while l < r:
                    m = (l + r) // 2
                    if potency > multipliers[m][1]:
                        l = m + 1
                    else:
                        r = m
                l -= 1
                multi = (multi * multipliers[l][0]) % MODULO
                potency -= multipliers[l][1]
            return (multi * multiplier) % MODULO

        return [(num * getMultiPotency(multiplier, potencies[i])) % MODULO for i, num in enumerate(nums)]
20700