<p>You are given a <strong>0-indexed</strong> <code>m x n</code> matrix <code>grid</code> consisting of <strong>positive</strong> integers.</p>

<p>You can start at <strong>any</strong> cell in the first column of the matrix, and traverse the grid in the following way:</p>

<ul>
	<li>From a cell <code>(row, col)</code>, you can move to any of the cells: <code>(row - 1, col + 1)</code>, <code>(row, col + 1)</code> and <code>(row + 1, col + 1)</code> such that the value of the cell you move to, should be <strong>strictly</strong> bigger than the value of the current cell.</li>
</ul>

<p>Return <em>the <strong>maximum</strong> number of <strong>moves</strong> that you can perform.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/04/11/yetgriddrawio-10.png" style="width: 201px; height: 201px;" />
<pre>
<strong>Input:</strong> grid = [[2,4,3,5],[5,4,9,3],[3,4,2,11],[10,9,13,15]]
<strong>Output:</strong> 3
<strong>Explanation:</strong> We can start at the cell (0, 0) and make the following moves:
- (0, 0) -&gt; (0, 1).
- (0, 1) -&gt; (1, 2).
- (1, 2) -&gt; (2, 3).
It can be shown that it is the maximum number of moves that can be made.</pre>

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

<pre>
<img alt="" src="https://assets.leetcode.com/uploads/2023/04/12/yetgrid4drawio.png" />
<strong>Input:</strong> grid = [[3,2,4],[2,1,9],[1,1,7]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> Starting from any cell in the first column we cannot perform any moves.
</pre>

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

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>2 &lt;= m, n &lt;= 1000</code></li>
	<li><code>4 &lt;= m * n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= grid[i][j] &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution:
    def maxMoves(self, grid: List[List[int]]) -> int:
         
        for r in grid:
            print(r)
        print('--')


        def rec(r,c,prev):
            
            if r >= len(grid) or r < 0 or c >= len(grid[0])  or prev >= grid[r][c]: 
                return 0

            if grid[r][c] > 0:
                grid[r][c] = - (1+ max(rec(r,c+1,grid[r][c] ),rec(r+1,c+1,grid[r][c] ),rec(r-1,c+1,grid[r][c] )))

            return abs(grid[r][c])
        
        _max = -1
        for i in range(len(grid)):
            _max = max(_max,rec(i,0,-1))
     
        return _max -1
1014