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 code384A {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner get = new Scanner(System.in);
		int n = get.nextInt();
		String[] s = new String[n];
		for(int i=0;i<n;i++){
			s[i]="";
		}
		int ans = 0;
		if(n==1){
			System.out.println(1);
			System.out.println("C");
			return;
		}
		String str="";
		for(int i=0;i<=n;i++){
			if(i%2==0){
				str=str+'C';
			}
			else{
				str=str+'.';
			}
		}
		String s1=str.substring(0, n);;
		String s2=str.substring(1, n+1);
		for(int i=0;i<n;i++){
			if(i%2==0){
				s[i]=s1;
				
			}
			else{
				s[i]=s2;
			}
		}
		if(n%2==0){
			ans=n*n/2;
		}
		else{
			ans=(int)(n*n/2)+1;
		}
		System.out.println(ans);
		for(int i=0;i<n;i++){
			System.out.println(s[i]);
		}
	}

}
