<p>You are given a <code>m x n</code> 2D array <code>board</code> representing a chessboard, where <code>board[i][j]</code> represents the <strong>value</strong> of the cell <code>(i, j)</code>.</p>

<p>Rooks in the <strong>same</strong> row or column <strong>attack</strong> each other. You need to place <em>three</em> rooks on the chessboard such that the rooks <strong>do not</strong> <strong>attack</strong> each other.</p>

<p>Return the <strong>maximum</strong> sum of the cell <strong>values</strong> on which the rooks are placed.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">board = </span>[[-3,1,1,1],[-3,1,-3,1],[-3,2,1,1]]</p>

<p><strong>Output:</strong> 4</p>

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

<p><img alt="" src="https://assets.leetcode.com/uploads/2024/08/08/rooks2.png" style="width: 294px; height: 450px;" /></p>

<p>We can place the rooks in the cells <code>(0, 2)</code>, <code>(1, 3)</code>, and <code>(2, 1)</code> for a sum of <code>1 + 1 + 2 = 4</code>.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">board = [[1,2,3],[4,5,6],[7,8,9]]</span></p>

<p><strong>Output:</strong> <span class="example-io">15</span></p>

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

<p>We can place the rooks in the cells <code>(0, 0)</code>, <code>(1, 1)</code>, and <code>(2, 2)</code> for a sum of <code>1 + 5 + 9 = 15</code>.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">board = [[1,1,1],[1,1,1],[1,1,1]]</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

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

<p>We can place the rooks in the cells <code>(0, 2)</code>, <code>(1, 1)</code>, and <code>(2, 0)</code> for a sum of <code>1 + 1 + 1 = 3</code>.</p>
</div>

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

<ul>
	<li><code>3 &lt;= m == board.length &lt;= 500</code></li>
	<li><code>3 &lt;= n == board[i].length &lt;= 500</code></li>
	<li><code>-10<sup>9</sup> &lt;= board[i][j] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def maximumValueSum(self, board: List[List[int]]) -> int:
        n=len(board)
        val=[]
        for idx,row in enumerate(board):
            for i,x in enumerate(row):
                row[i]=(x,i,idx)
            val.extend(sorted(row,reverse=True)[:3])
        val.sort(reverse=True)
        m=len(val)
        adj=[0 for _ in range(m)]
        for i in range(m-1):
            for j in range(i+1,m):
                if val[i][1]!=val[j][1] and val[i][2]!=val[j][2]:
                    adj[i]|=1<<j
                    adj[j]|=1<<i
        res=-inf
        
        def find(n):
            res=0
            while not n&((1<<64)-1):
                n>>=64
                res+=64
            if not n&((1<<32)-1):
                n>>=32
                res+=32
            if not n&((1<<16)-1):
                n>>=16
                res+=16
            if not n&((1<<8)-1):
                n>>=8
                res+=8
            if not n&((1<<4)-1):
                n>>=4
                res+=4
            if not n&((1<<2)-1):
                n>>=2
                res+=2
            if not n&((1<<1)-1):
                n>>=1
                res+=1
            return res
        
        for i in range(m-1):
            for j in range(i+1,m):
                if not adj[i]&(1<<j):
                    continue
                s=adj[i]&adj[j]
                if s:
                    idx=find(s)
                    res=max(res,val[i][0]+val[j][0]+val[idx][0])
        return res
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