#include <bits/stdc++.h>
#pragma warning(disable : 4996)
int lol(const void *x1, const void *x2) { return *(int *)x1 - *(int *)x2; }
int main(void) {
  int n, i, j;
  char c0[1001], c1[1001];
  scanf("%i", &n);
  if (n % 2 == 0)
    printf("%i\n", (n * n) / 2);
  else
    printf("%i\n", (n * n) / 2 + 1);
  for (i = 0; i < 2; i++) {
    for (j = 0; j < n; j++) {
      if ((i + j) % 2 == 0) {
        if (i == 0)
          c0[j] = 'C';
        else
          c1[j] = 'C';
      } else {
        if (i == 0)
          c0[j] = '.';
        else
          c1[j] = '.';
      }
    }
    c0[j] = '\0';
    c1[j] = '\0';
  }
  for (i = 0; i < n; i++) {
    if (i % 2 == 0)
      puts(c0);
    else
      puts(c1);
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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

### Response

```cpp
#include <bits/stdc++.h>
#pragma warning(disable : 4996)
int lol(const void *x1, const void *x2) { return *(int *)x1 - *(int *)x2; }
int main(void) {
  int n, i, j;
  char c0[1001], c1[1001];
  scanf("%i", &n);
  if (n % 2 == 0)
    printf("%i\n", (n * n) / 2);
  else
    printf("%i\n", (n * n) / 2 + 1);
  for (i = 0; i < 2; i++) {
    for (j = 0; j < n; j++) {
      if ((i + j) % 2 == 0) {
        if (i == 0)
          c0[j] = 'C';
        else
          c1[j] = 'C';
      } else {
        if (i == 0)
          c0[j] = '.';
        else
          c1[j] = '.';
      }
    }
    c0[j] = '\0';
    c1[j] = '\0';
  }
  for (i = 0; i < n; i++) {
    if (i % 2 == 0)
      puts(c0);
    else
      puts(c1);
  }
  return 0;
}

```