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.Scanner;

/**
 * Created by yuu on 24/3/17.
 */
public class Problem384A {
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        StringBuilder odd = new StringBuilder();
        int oddC = (int)Math.ceil((double)n/2);
        int evenC = (int)Math.floor((double)n/2);
        for (int i = 0; i < n; i++){
            if (i % 2 == 0) odd.append('C');
            else odd.append('.');
        }
        StringBuilder even = new StringBuilder(odd);
        if (even.charAt(even.length()-1) == 'C')
            even.append('.');
        else
            even.append('C');
        even.deleteCharAt(0);

        int total = oddC * oddC + evenC * evenC;
        System.out.println(total);
        for (int i = 1; i <= n; i++){
            if (i % 2 == 0) {
                System.out.println(even);
            }
            else {
                System.out.println(odd);
            }
        }

    }
}
