<p>You are given an <code>m x n</code> matrix <code>mat</code> that has its rows sorted in non-decreasing order and an integer <code>k</code>.</p>

<p>You are allowed to choose <strong>exactly one element</strong> from each row to form an array.</p>

<p>Return <em>the </em><code>k<sup>th</sup></code><em> smallest array sum among all possible arrays</em>.</p>

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

<pre>
<strong>Input:</strong> mat = [[1,3,11],[2,4,6]], k = 5
<strong>Output:</strong> 7
<strong>Explanation:</strong> Choosing one element from each row, the first k smallest sum are:
[1,2], [1,4], [3,2], [3,4], [1,6]. Where the 5th sum is 7.
</pre>

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

<pre>
<strong>Input:</strong> mat = [[1,3,11],[2,4,6]], k = 9
<strong>Output:</strong> 17
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> mat = [[1,10,10],[1,4,5],[2,3,6]], k = 7
<strong>Output:</strong> 9
<strong>Explanation:</strong> Choosing one element from each row, the first k smallest sum are:
[1,1,2], [1,1,3], [1,4,2], [1,4,3], [1,1,6], [1,5,2], [1,5,3]. Where the 7th sum is 9.  
</pre>

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

<ul>
	<li><code>m == mat.length</code></li>
	<li><code>n == mat.length[i]</code></li>
	<li><code>1 &lt;= m, n &lt;= 40</code></li>
	<li><code>1 &lt;= mat[i][j] &lt;= 5000</code></li>
	<li><code>1 &lt;= k &lt;= min(200, n<sup>m</sup>)</code></li>
	<li><code>mat[i]</code> is a non-decreasing array.</li>
</ul>

class Solution:
    def kthSmallest(self, mat: List[List[int]], k: int) -> int:
        M, N = len(mat), len(mat[0])

        initial_value = sum(row[0] for row in mat)
    
        first_candidate = (initial_value,) + (0,) * M
        visited_candidates = {first_candidate}
        candidates = [first_candidate]

        for _ in range(k - 1):
            candidate = list(heapq.heappop(candidates))
            candidate_value, candidate_index_layout = candidate[0], candidate[1:]

            for row_index, column_index in enumerate(candidate_index_layout):
                if column_index + 1 < N:
                    next_candidate_value = candidate_value + mat[row_index][column_index + 1] - mat[row_index][column_index]
                    candidate_index_layout[row_index] += 1
                    
                    next_candidate = tuple([next_candidate_value] + candidate_index_layout)
                    if next_candidate not in visited_candidates:
                        heapq.heappush(candidates, next_candidate)
                        visited_candidates.add(next_candidate)

                    candidate_index_layout[row_index] -= 1

        candidate = list(heapq.heappop(candidates))
        return candidate[0]
20200