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 namespace std;
int arr[1005][1005];
void pre() {
  for (__typeof(1005) i = 0; i < (1005); i++) {
    for (__typeof(1005) j = 0; j < (1005); j++) {
      arr[i][j] = 0;
    }
  }
}
bool isSafe(int i, int j, int N) {
  if (i < 0 || j < 0 || i >= N || j >= N)
    return false;
  else if (arr[i][j])
    return false;
  return true;
}
void doThis() {}
int main() {
  pre();
  int x[] = {1, -1, 0, 0};
  int y[] = {0, 0, 1, -1};
  int N;
  cin >> N;
  long long ans = 0;
  for (__typeof(N) i = 0; i < (N); i++) {
    for (__typeof(N) j = 0; j < (N); j++) {
      if (arr[i][j] == 0) {
        arr[i][j] = 1;
        for (__typeof(4) m = 0; m < (4); m++) {
          if (isSafe(i + x[m], j + y[m], N)) arr[i + x[m]][j + y[m]] = -1;
        }
        ans++;
      }
    }
  }
  cout << ans;
  printf("\n");
  for (__typeof(N) i = 0; i < (N); i++) {
    for (__typeof(N) j = 0; j < (N); j++) {
      if (arr[i][j] == 1)
        printf("C");
      else
        printf(".");
    }
    printf("\n");
  }
  return 0;
}
