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.util.*;
import java.io.*;
public class Main{
  public static void main(String []args)throws IOException{
    BufferedReader br  = new BufferedReader(new InputStreamReader(System.in));
    PrintWriter pw = new PrintWriter(new OutputStreamWriter(System.out));
    StringTokenizer st ;
    int n = Integer.parseInt(br.readLine());
    if(n%2==0)
    pw.println(n*n/2);
    else
      pw.println(((int)Math.ceil((n*1.0)/2)*n)-n/2);
    char k = 'C' ;
    char kx = '.';
    String s = "";
    if(n%2==0){
    for(int j = 1 ; j <=n ; j++){
      for(int i = 1 ; i <= n ; i++){
       if(i%2==0)
         s+=kx;
       else
         s+=k;
      }
      pw.println(s);
      if(s.charAt(n-1)=='C'){
        k = 'C';
        kx = '.';
      }
      else{
        k = '.';
        kx = 'C';
      }
      s="";
    } 
    }
    else{
    for(int j = 1 ; j <=n ; j++){
      for(int i = 1 ; i <= n ; i++){
       if(i%2==0)
         s+=kx;
       else
         s+=k;
      }
      pw.println(s);
      if(s.charAt(n-1)=='C'){
        k = '.';
        kx = 'C';
      }
      else{
        k = 'C';
        kx = '.';
      }
      s="";
    }
    }
    
    pw.close();
  }
}
