Given an `m x n` grid. Each cell of the grid has a sign pointing to the next cell you should visit if you are currently in this cell. The sign of `grid[i][j]` can be:

- `1` which means go to the cell to the right. (i.e go from `grid[i][j]` to `grid[i][j + 1]`)
- `2` which means go to the cell to the left. (i.e go from `grid[i][j]` to `grid[i][j - 1]`)
- `3` which means go to the lower cell. (i.e go from `grid[i][j]` to `grid[i + 1][j]`)
- `4` which means go to the upper cell. (i.e go from `grid[i][j]` to `grid[i - 1][j]`)

Notice that there could be some signs on the cells of the grid that point outside the grid.

You will initially start at the upper left cell `(0, 0)`. A valid path in the grid is a path that starts from the upper left cell `(0, 0)` and ends at the bottom-right cell `(m - 1, n - 1)` following the signs on the grid. The valid path does not have to be the shortest.

You can modify the sign on a cell with `cost = 1`. You can modify the sign on a cell one time only.

Return the minimum cost to make the grid have at least one valid path.

Example 1:

    Input: grid = [[1,1,1,1],[2,2,2,2],[1,1,1,1],[2,2,2,2]]
    Output: 3
    Explanation: You will start at point (0, 0).
    The path to (3, 3) is as follows. (0, 0) --> (0, 1) --> (0, 2) --> (0, 3) change the arrow to down with cost = 1 --> (1, 3) --> (1, 2) --> (1, 1) --> (1, 0) change the arrow to down with cost = 1 --> (2, 0) --> (2, 1) --> (2, 2) --> (2, 3) change the arrow to down with cost = 1 --> (3, 3)
    The total cost = 3.

Example 2:

    Input: grid = [[1,1,3],[3,2,2],[1,1,4]]
    Output: 0
    Explanation: You can follow the path from (0, 0) to (2, 2).

Example 3:

    Input: grid = [[1,2],[4,3]]
    Output: 1

Constraints:

- `m == grid.length`
- `n == grid[i].length`
- `1 <= m, n <= 100`
- `1 <= grid[i][j] <= 4`