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.Arrays;
import java.util.Scanner;

public class Coder {

	static int dx[] = { 1, 0, -1, 0 };
	static int dy[] = { 0, 1, 0, -1 };
	static int maxr, maxc;

	public static boolean isValid(int r, int c) {
		return r >= 0 && r < maxr && c >= 0 && c < maxc;
	}

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = maxr = maxc = sc.nextInt();
		int c = 0;
		char [][] grid = new char[n][n];
		for(int i = 0;i < n;i++){
			for(int j = 0;j < n;j++){
				boolean f = true;
				for(int k = 0;k < dx.length;k++){
					int newr = i + dx[k];
					int newc = j + dy[k];
					if(isValid(newr, newc)){
						if(grid[newr][newc] == 'C')
							f = false;
					}
				}
				if(f){
					grid[i][j] = 'C';
					c++;
				}
				else
					grid[i][j] = '.';
			}
		}
		System.out.println(c);
		for(int i = 0;i < n;i++)
			System.out.println(new String(grid[i]));
	}

}
