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 java.io.*;
public class Main{
    static String drawEvenRow(int n,char[] a){
        for(int i=0;i<n;i++){
            if(i%2==0)
                a[i]='C';
            else
                a[i]='.';
        }
         return new String(a);
    }
    static String drawOddRow(int n,char[] a){
        for(int i=0;i<n;i++){
            if(i%2==0)
                a[i]='.';
            else
                a[i]='C';
        }
         return new String(a);
    }
    public static void main(String[] args) throws IOException{
        int n;
        BufferedReader input=new BufferedReader(new InputStreamReader(System.in));
        n=Integer.parseInt(input.readLine());
        char[] s1=new char[n];
        char[] s2=new char[n];
        
        String evenRow=drawEvenRow(n,s1);
        String oddRow=drawOddRow(n,s2);
        
        if(n%2==0)
            System.out.println(n*n/2);
        else
            System.out.println((n*n+1)/2);
        
        for(int i=0;i<n;i++){
            if(i%2==0)
                System.out.println(evenRow);
            else
                System.out.println(oddRow);
        }
                
    }
    
}
