mat = [
    ['.','o','.','.','.'],
    ['.','.','.','o','.'],
    ['.','.','.','.','o'],
    ['.','o','.','.','.'],
    ['.','.','.','o','.']
]

#Size of given matrix is N X N
N = 5


# A utility function to find minimum
# of two numbers
def getMin(x, y):
    if x < y:
        return x
    else:
        return y

# Returns size of Maximum size
# subsquare matrix surrounded by 'X'
def findSubSquare(mat):

    #####   Create helper arrays    #####
    #####################################

    Max = 0 # Initialize results
    Bx = 0
    By = 0


    # Initialize the left-top value
    # in hor[][] and ver[][]
    hor = [[0 for i in range(N)]
              for i in range(N)]
    ver = [[0 for i in range(N)]
              for i in range(N)]

    if mat[0][0] == '.':
        hor[0][0] = 1
        ver[0][0] = 1

    #print('hor: ' + str(hor))
    #print('ver: ' + str(ver))

    # Fill values in hor[][] and ver[][]
    for i in range(N):

        for j in range(N):

            if (mat[i][j] == 'o'):
                ver[i][j] = 0
                hor[i][j] = 0

            else:
                # if j == 0 or i == 0:
                #     ver[i][j] = 1
                #     hor[i][j] = 1
                #
                # #elif i == 0:
                # #    ver[i][j] = 1
                #
                # else:
                if i == 0:
                    ver[i][j] = 1
                else:
                    ver[i][j] = ver[i - 1][j] + 1

                if j == 0:
                    hor[i][j] = 1
                else:
                    hor[i][j] = hor[i][j - 1] + 1

    #print('hor: ' + str(hor))
    #print('ver: ' + str(ver))




    #####   Find sub squares    #####
    #################################



    # Start from the rightmost-bottommost corner
    # element and find the largest ssubsquare
    # with the help of hor[][] and ver[][]
    for i in range(N - 1, 0, -1):

        for j in range(N - 1, 0, -1):

            # Find smaller of values in hor[][] and
            # ver[][]. A Square can only be made by
            # taking smaller value
            small = getMin(hor[i][j], ver[i][j])
            #print(str(small) + ': (' + str(j) + ',' + str(i) + ')')

            # At this point, we are sure that there
            # is a right vertical line and bottom
            # horizontal line of length at least 'small'.

            # We found a bigger square if following
            # conditions are met:
            # 1)If side of square is greater than Max.
            # 2)There is a left vertical line
            #   of length >= 'small'
            # 3)There is a top horizontal line
            #   of length >= 'small'
            while (small > Max):

                if (ver[i][j - small + 1] >= small and hor[i - small + 1][j] >= small):

                    Max = small
                    Bx = j
                    By = i

                small -= 1

    #print(Bx, By)
    return (Max, Bx, By)

m, x, y = findSubSquare(mat)
print(f'>> ({x}, {y}) <<')
#print(By)
for i, row in enumerate(mat):
    for j, ch in enumerate(row):
        if (i <= y and i > (y-m)) and (j <= x and j > (x-m)):
            print ('x', end = ' ')
        else:
            print(ch, end=' ')
    print('')