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 evenGrid(n):
    result = list()
    cur1 = ""
    cur2 = ""
    for i in range(0,n,2):
        cur1+="C."
    for i in range(0,n,2):
        cur2+=".C"
    for i in range(0,n,2):
        result.append(cur1)
        result.append(cur2)

    return result

# build the grid when n is odd
def oddGrid(n):
    result = list()
    cur1 = ""
    cur2 = ""
    for i in range(0,n,2):
        cur1+="C."
    cur1 = cur1[:-1]
    for i in range(0,n,2):
        cur2+=".C"
    cur2 = cur2[:-1]
    for i in range(0,n,2):
        result.append(cur1)
        result.append(cur2)
    del result[-1]
    return result


# calculate the total pieces based on n being odd or not
def solve(n):
    result = ()
    if(n%2==0):
        result = evenGrid(n)
        print (n/2) * n
    else:
        result = oddGrid(n)
        print ((n/2)+1)*((n/2)+1) + (n/2)*(n/2)

    for i in range(0,len(result)):
        print result[i]

if __name__ == "__main__":
    n = int(raw_input())
    solve(n)