On a 2-dimensional `grid`

, there are 4 types of squares:

`1`

represents the starting square. There is exactly one starting square.`2`

represents the ending square. There is exactly one ending square.`0`

represents empty squares we can walk over.`-1`

represents obstacles that we cannot walk over.

Return the number of 4-directional walks from the starting square to the ending square, that **walk over every non-obstacle square exactly once**.

**Example 1:**

Input:[[1,0,0,0],[0,0,0,0],[0,0,2,-1]]Output:2Explanation:We have the following two paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2) 2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)

**Example 2:**

Input:[[1,0,0,0],[0,0,0,0],[0,0,0,2]]Output:4Explanation:We have the following four paths: 1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3) 2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3) 3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3) 4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)

**Example 3:**

Input:[[0,1],[2,0]]Output:0Explanation:There is no path that walks over every empty square exactly once. Note that the starting and ending square can be anywhere in the grid.

**Note:**

`1 <= grid.length * grid[0].length <= 20`

class Solution {
public int uniquePathsIII(int[][] grid) {
}
}