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
import math,string,itertools,fractions,heapq,collections,re,array,bisect,sys,random,time,queue,copy

sys.setrecursionlimit(10**7)
inf=10**20
mod=10**9+7
dd=[(-1,0),(0,1),(1,0),(0,-1)]
ddn=[(-1,0),(-1,1),(0,1),(1,1),(1,0),(1,-1),(0,-1),(-1,-1)]

def LI(): return [int(x) for x in sys.stdin.readline().split()]
def LI_(): return [int(x)-1 for x in sys.stdin.readline().split()]
def I(): return int(sys.stdin.readline())
def LS(): return sys.stdin.readline().split()
def S(): return input()

def main():
  n=I()
  l=[['.']*n for _ in range(n)]

  if n==1:
    print(1)
    print('C')

  else:

    sm1=sm2=0
    for i in range(2):
      for j in range(n):
        f=True
        for y,x in dd:
          if i+y<0 or j+x<0 or i+y>=n or j+x>=n:
            continue
  
          if l[i+y][j+x]=='C':
            f=False
  
        if f:
          l[i][j]='C'
          if i==0:
            sm1+=1
          else:
            sm2+=1
  
    # print(sm1,sm2,(-(-n//2)),n//2)
    print(sm1*(-(-n//2))+sm2*(n//2))
    for i in range(n):
      if i%2==0:
        x=l[0]
      else:
        x=l[1]
      print(''.join(x))

main()
# print(main())
