Iahub likes chess very much. He even invented a new chess piece named Coder. A Coder can move (and attack) one square horizontally or vertically. More precisely, if the Coder is located at position (x, y), he can move to (or attack) positions (x + 1, y), (x–1, y), (x, y + 1) and (x, y–1).

Iahub wants to know how many Coders can be placed on an n × n chessboard, so that no Coder attacks any other Coder.

Input

The first line contains an integer n (1 ≤ n ≤ 1000).

Output

On the first line print an integer, the maximum number of Coders that can be placed on the chessboard.

On each of the next n lines print n characters, describing the configuration of the Coders. For an empty cell print an '.', and for a Coder print a 'C'.

If there are multiple correct answers, you can print any.

Examples

Input

2


Output

2
C.
.C
def int_lst_input():
    return [int(val) for val in input().split()]


def int_input():
    return int(input())


def print_lst(lst):
    print(' '.join([str(val) for val in lst]))


def solve():
    n = int_input()

    grid = []

    even = ''.join(['C' if i % 2 == 0 else '.' for i in range(n)])
    odd = ''.join(['C' if i % 2 == 1 else '.' for i in range(n)])

    c_even_count = sum([1 if c == 'C' else 0 for c in even])
    c_odd_count = sum([1 if c == 'C' else 0 for c in odd])

    c = c_even_count * (n // 2 + n % 2) + c_odd_count * (n // 2)

    for i in range(n):
        if i % 2 == 0:
            grid.append(even)
        else:
            grid.append(odd)

    print(c)
    for row in grid:
        print(row)


if __name__ == '__main__':
    solve()
