Given a `m * n`

matrix of ones and zeros, return how many **square** submatrices have all ones.

## Example 1:

```
Input: matrix =
[
[0,1,1,1],
[1,1,1,1],
[0,1,1,1]
]
Output: 15
Explanation:
There are 10 squares of side 1.
There are 4 squares of side 2.
There is 1 square of side 3.
Total number of squares = 10 + 4 + 1 = 15.
```

## Example 2:

```
Input: matrix =
[
[1,0,1],
[1,1,0],
[1,1,0]
]
Output: 7
Explanation:
There are 6 squares of side 1.
There is 1 square of side 2.
Total number of squares = 6 + 1 = 7.
```

**Constraints:**

```
1 <= arr.length <= 300
1 <= arr[0].length <= 300
0 <= arr[i][j] <= 1
```

## Solution:

```
class Solution:
def countSquares(matrix):
p_arr = [[0 for i in range(len(matrix[0]))] for j in range(len(matrix))]
result = 0
for index_i in range(1, len(matrix)):
for index_j in range(1, len(matrix[0])):
if matrix[index_i][index_j] == 1:
temp_mdata = matrix[index_i-1][index_j], matrix[index_i][index_j-1]
matrix[index_i][index_j] = min(matrix[index_i-1][index_j-1], min(temp_mdata))+1
return sum([ sum(x) for x in matrix])
input = [[0,1,1,1],[1,1,1,1],[0,1,1,1]]
print(Solution.countSquares(input))
input = [[1,0,1],[1,1,0],[1,1,0]]
print(Solution.countSquares(input))
```