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
#include <bits/stdc++.h>
using pii = std::pair<long long, long long>;
using namespace std;
long long n, grid[1005][1005] = {0};
class ACoder {
 public:
  void solve(std::istream& in, std::ostream& out) {
    in >> n;
    long long cnt = 0;
    for (long long i = (0); i < (n); i++)
      for (long long j = (0); j < (n); j++)
        if ((!i || !grid[i - 1][j]) && (!j || !grid[i][j - 1])) {
          grid[i][j] = 1;
          cnt++;
        }
    char ans[2] = {'.', 'C'};
    out << cnt << "\n";
    for (long long i = (0); i < (n); i++) {
      for (long long j = (0); j < (n); j++) out << ans[grid[i][j]];
      out << "\n";
    }
  }
};
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  ACoder solver;
  std::istream& in(std::cin);
  std::ostream& out(std::cout);
  solver.solve(in, out);
  return 0;
}
