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;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskA solver = new TaskA();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskA {
    public void solve(int testNumber, Scanner in, PrintWriter out) {
		int n = in.nextInt();
		int offset = 0;
		int counter = 0;

		if (n%2 == 0) out.println(n*n/2);
		else out.println((n * n / 2) + 1);

		for (int i = 0; i < n; i++) {
			for (int j = 0; j < n; j++) {
				if (offset == 0){
					if (j % 2 == 0) out.print('C');
					else out.print('.');
				}
				if (offset == 1){
					if (j%2 != 0) out.print('C');
					else out.print('.');
				}
			}
			if (offset == 0) offset = 1;
			else offset = 0;
			out.println();
		}
	}
}

