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[]) {

    Scanner input = new Scanner(System.in);
    StringBuilder sb = new StringBuilder();
    
    int n = input.nextInt();
    char grid[][] = new char[n][n];
     
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            grid[i][j] = '.';
        }
    }
     
    int dx[] = { 1, -1, 0, 0  },
        dy[] = { 0, 0, 1, -1  },
        res = 0;
    
    for (int i = 0; i < n; i++) {
        sb.append('\n');
         
        for (int j = 0; j < n; j++) {
            if (grid[i][j] == '.') {
                grid[i][j] = 'C';
                res++;
                sb.append('C');
                 
                for (int k = 0; k < 4; k++) {
                    if (i + dx[k] < n && i + dx[k] >= 0 && j + dy[k] < n && j + dy[k] >= 0) {
                        grid[i + dx[k]][j + dy[k]] = 'O';
                    }
                }
            } else {
                sb.append('.');
            }
        }
    }
     
    System.out.print(res + sb.toString());
             
    input.close();
         
    }
 
}