Given a picture consisting of black and white pixels, and a positive integer N, find the number of black pixels located at some specific row **R** and column **C** that align with all the following rules:

- Row R and column C both contain exactly N black pixels.
- For all rows that have a black pixel at column C, they should be exactly the same as row R

The picture is represented by a 2D char array consisting of 'B' and 'W', which means black and white pixels respectively.

**Example:**

Input:[['W', 'B', 'W', 'B', 'B', 'W'], ['W', 'B', 'W', 'B', 'B', 'W'], ['W', 'B', 'W', 'B', 'B', 'W'], ['W', 'W', 'B', 'W', 'B', 'W']] N = 3Output:6Explanation:All the bold 'B' are the black pixels we need (all 'B's at column 1 and 3). 0 1 2 3 4 5 column index 0 [['W','B', 'W','B', 'B', 'W'], 1 ['W','B', 'W','B', 'B', 'W'], 2 ['W','B', 'W','B', 'B', 'W'], 3 ['W', 'W', 'B', 'W', 'B', 'W']] row index Take 'B' at row R = 0 and column C = 1 as an example: Rule 1, row R = 0 and column C = 1 both have exactly N = 3 black pixels. Rule 2, the rows have black pixel at column C = 1 are row 0, row 1 and row 2. They are exactly the same as row R = 0.

**Note:**

- The range of width and height of the input 2D array is [1,200].

```
struct Solution;
impl Solution {
fn find_black_pixel(picture: Vec<Vec<char>>, n: i32) -> i32 {
let k = n as usize;
let n = picture.len();
let m = picture[0].len();
let mut rows: Vec<Vec<usize>> = vec![vec![]; n];
let mut cols: Vec<Vec<usize>> = vec![vec![]; m];
for i in 0..n {
for j in 0..m {
if picture[i][j] == 'B' {
rows[i].push(j);
cols[j].push(i);
}
}
}
let mut res = 0;
for i in 0..n {
for j in 0..m {
if picture[i][j] == 'B' && rows[i].len() == k && cols[j].len() == k {
if cols[j].iter().all(|&r| rows[r] == rows[i]) {
res += 1;
}
}
}
}
res
}
}
#[test]
fn test() {
let picture = vec_vec_char![
['W', 'B', 'W', 'B', 'B', 'W'],
['W', 'B', 'W', 'B', 'B', 'W'],
['W', 'B', 'W', 'B', 'B', 'W'],
['W', 'W', 'B', 'W', 'B', 'W']
];
let n = 3;
let res = 6;
assert_eq!(Solution::find_black_pixel(picture, n), res);
}
```