<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 {
public:
    int kthLargestValue(vector<vector<int>>& matrix, int k) {
        int m = matrix.size(), n = matrix[0].size(), i, j;
        vector<vector<int>> xx(m, vector<int>(n));
        vector<int> results;
        for (i = 0; i < m; i++) {
            for (j = 0; j < n; j++) {
                if (i == 0 && j == 0) xx[0][0] = matrix[0][0];
                else if (i == 0) xx[0][j] = xx[0][j-1] ^ matrix[0][j];
                else if (j == 0) xx[i][0] = xx[i-1][0] ^ matrix[i][0];
                else xx[i][j] = xx[i-1][j-1] ^ xx[i-1][j] ^ xx[i][j-1] ^ matrix[i][j];
                results.push_back(xx[i][j]);
            }
        }
        ranges::nth_element(results, results.end() - k);
        return results[results.size() - k];
        // nth_element(results.begin(), results.end() - k, results.end());
        // return results[results.size() - k];
        // nth_element(results.begin(), results.begin() + k - 1, results.end(), greater<int>());
        // return results[k - 1];
    }
};
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