The goal of 8 Queens Problem is to put eight queens on a chess-board such that none of them threatens any of others. A queen threatens the squares in the same row, in the same column, or on the same diagonals as shown in the following figure.

<image>


For a given chess board where $k$ queens are already placed, find the solution of the 8 queens problem.

Constraints

* There is exactly one solution

Input

In the first line, an integer $k$ is given. In the following $k$ lines, each square where a queen is already placed is given by two integers $r$ and $c$. $r$ and $c$ respectively denotes the row number and the column number. The row/column numbers start with 0.

Output

Print a $8 \times 8$ chess board by strings where a square with a queen is represented by 'Q' and an empty square is represented by '.'.

Example

Input

2
2 2
5 3


Output

......Q.
Q.......
..Q.....
.......Q
.....Q..
...Q....
.Q......
....Q...
#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (int)(n); i++)
using namespace std;

int main() {
   int k;
   cin >> k;
   vector<string> data(8,"........");
   vector<int> perm(8);
   rep(i,8) perm[i] = i;
   vector<int> x(k), y(k);
   rep(i,k) {
      int X, Y;
      cin >> X >> Y;
      x[i] = X;
      y[i] = Y;
   }
   do {
      bool ok = true;
      rep(i,k) {
         if (perm[y[i]] != x[i]) ok = false;
      }
      if (ok) {
         bool ans = true;
         rep(i,7) {
            for (int j = 1; i+j < 8; j++) {
               if (perm[i] + j == perm[i+j]) ans = false;
               if (perm[i] - j == perm[i+j]) ans = false;
            }
         }
         if (ans) {
            rep(i,8) {
               data[perm[i]][i] = 'Q';
            }
            rep(i,8) {
               rep(j,8) {
                  cout << data[i][j];
               }
               cout << endl;
            }
            return 0;
         }
      }
   } while(next_permutation(perm.begin(),perm.end()));
}
