<p>You are given a 2D <code>matrix</code> of size <code>m x n</code>, consisting of non-negative integers. You are also given an integer <code>k</code>.</p>

<p>The <strong>value</strong> of coordinate <code>(a, b)</code> of the matrix is the XOR of all <code>matrix[i][j]</code> where <code>0 &lt;= i &lt;= a &lt; m</code> and <code>0 &lt;= j &lt;= b &lt; n</code> <strong>(0-indexed)</strong>.</p>

<p>Find the <code>k<sup>th</sup></code> largest value <strong>(1-indexed)</strong> of all the coordinates of <code>matrix</code>.</p>

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

<pre>
<strong>Input:</strong> matrix = [[5,2],[1,6]], k = 1
<strong>Output:</strong> 7
<strong>Explanation:</strong> The value of coordinate (0,1) is 5 XOR 2 = 7, which is the largest value.
</pre>

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

<pre>
<strong>Input:</strong> matrix = [[5,2],[1,6]], k = 2
<strong>Output:</strong> 5
<strong>Explanation:</strong> The value of coordinate (0,0) is 5 = 5, which is the 2nd largest value.
</pre>

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

<pre>
<strong>Input:</strong> matrix = [[5,2],[1,6]], k = 3
<strong>Output:</strong> 4
<strong>Explanation:</strong> The value of coordinate (1,0) is 5 XOR 1 = 4, which is the 3rd largest value.</pre>

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

<ul>
	<li><code>m == matrix.length</code></li>
	<li><code>n == matrix[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 1000</code></li>
	<li><code>0 &lt;= matrix[i][j] &lt;= 10<sup>6</sup></code></li>
	<li><code>1 &lt;= k &lt;= m * n</code></li>
</ul>

class Solution:
    def kthLargestValue(self, matrix: List[List[int]], k: int) -> int:
        xors = self.construct_xors(matrix)
        return self.quick_select(xors, k)

    
    def construct_xors(self, xors):
        m,n = len(xors),len(xors[0])

        def valid_coord(i,j):
            return 0 <= i < m and 0 <= j < n
        
        for i in range(m):
            for j in range(n):
                for di, dj in ((-1, 0), (0, -1), (-1, -1)):
                    ci, cj = i + di, j + dj
                    xors[i][j] ^= xors[ci][cj] * valid_coord(ci, cj)

        xors = [xor for row in xors for xor in row]
        return xors

    def quick_select(self, container, k):
        pivot = choice(container)

        g = [xor for xor in container if xor > pivot]
        e = [xor for xor in container if xor == pivot]
        l = [xor for xor in container if xor < pivot]

        if len(g) >= k:
            return self.quick_select(g, k)
        elif len(g) + len(e) >= k:
            return e[0]
        else:
            return self.quick_select(l, k - len(g) - len(e))
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