<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:
        m, n = len(matrix), len(matrix[0])

        # Initialize prefix_xor with the same dimensions as the matrix
        prefix_xor = [[0] * n for _ in range(m)]
        prefix_xor[0][0] = matrix[0][0]

        # Compute the prefix XOR values
        for i in range(m):
            for j in range(n):
                if i == 0 and j == 0:
                    continue
                if i > 0:
                    prefix_xor[i][j] ^= prefix_xor[i - 1][j]
                if j > 0:
                    prefix_xor[i][j] ^= prefix_xor[i][j - 1]
                if i > 0 and j > 0:
                    prefix_xor[i][j] ^= prefix_xor[i - 1][j - 1]
                prefix_xor[i][j] ^= matrix[i][j]

        # Collect all the values from prefix_xor
        result = []
        for row in prefix_xor:
            result.extend(row)

        # Sort the values in descending order and get the k-th largest value
        result.sort(reverse=True)
        return result[k - 1]
47000