#include <bits/stdc++.h>
using namespace std;
vector<int> v;
string int_to_string(int a) { return to_string(a); }
double string_to_int(string s) {
  stringstream x(s);
  double a = 0;
  x >> a;
  return a;
}
long long int gcd(long long int a, long long int b) {
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
int main() {
  int b, c, n, cnt = 0;
  cin >> n;
  char a[n][n];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if ((i % 2 == 0 && j % 2 == 0) || (i % 2 != 0 && j % 2 != 0)) {
        a[i][j] = 'C';
        cnt++;
      } else
        a[i][j] = '.';
    }
  }
  cout << cnt << endl;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cout << a[i][j];
    }
    cout << endl;
  }
}

### Prompt

Please formulate a cpp solution to the following problem:

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>
using namespace std;
vector<int> v;
string int_to_string(int a) { return to_string(a); }
double string_to_int(string s) {
  stringstream x(s);
  double a = 0;
  x >> a;
  return a;
}
long long int gcd(long long int a, long long int b) {
  if (a % b == 0) return b;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
int main() {
  int b, c, n, cnt = 0;
  cin >> n;
  char a[n][n];
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if ((i % 2 == 0 && j % 2 == 0) || (i % 2 != 0 && j % 2 != 0)) {
        a[i][j] = 'C';
        cnt++;
      } else
        a[i][j] = '.';
    }
  }
  cout << cnt << endl;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      cout << a[i][j];
    }
    cout << endl;
  }
}

```