CodeForces 384A Coder

Coder

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 384A

Description

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.

Sample Input

Input

2

Output

2
C.
.C

 #include <stdio.h>
 #include <string.h>
 int main()
 {
     int n;
     int i,j,k;
     while(scanf("%d",&n)!=EOF)
     {
         if(n%==)
         {
             printf("%d\n",n*n/);
             for(i=;i<=n/;i++)
             {
                 for(j=;j<=n;j++)
                     if(j%==)
                         printf("C");
                     else
                         printf(".");
                 printf("\n");
                 for(j=;j<=n;j++)
                     if(j%==)
                         printf("C");
                     else
                         printf(".");
                 printf("\n");
             }
         }
         else
         {
             printf("%d\n",(n+)*(n+)/+(n-)*(n-)/);
             for(i=;i<=n;i++)
             {
                 if(i%==)
                 {
                     for(j=;j<=n;j++)
                         if(j%==)
                             printf("C");
                         else
                             printf(".");
                     printf("\n");
                 }
                 else
                 {
                     for(j=;j<=n;j++)
                         if(j%==)
                             printf("C");
                         else
                             printf(".");
                     printf("\n");
                 }
             }
         }
     }
     return ;
 }