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%2==0)
        {
            printf("%d
",n*n/2);
            for(i=1;i<=n/2;i++)
            {
                for(j=1;j<=n;j++)
                    if(j%2==1)
                        printf("C");
                    else
                        printf(".");
                printf("
");
                for(j=1;j<=n;j++)
                    if(j%2==0)
                        printf("C");
                    else
                        printf(".");
                printf("
");
            }
        }
        else
        {
            printf("%d
",(n+1)*(n+1)/4+(n-1)*(n-1)/4);
            for(i=1;i<=n;i++)
            {
                if(i%2==1)
                {
                    for(j=1;j<=n;j++)
                        if(j%2==1)
                            printf("C");
                        else
                            printf(".");
                    printf("
");
                }
                else
                {
                    for(j=1;j<=n;j++)
                        if(j%2==0)
                            printf("C");
                        else
                            printf(".");
                    printf("
");
                }
            }
        }
    }
    return 0;
}