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>
       
using namespace std;
typedef long long ll;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

int main() {
  int k;
  cin >> k;

  vector<int> R(8, -1);
  for (int i = 0; i < (int)(k); ++i) {
    int r, c;
    cin >> r >> c;
    R[r] = c;
  }

  vector<int> perm(8);
  for (int i = 0; i < (int)(8); ++i) perm[i] = i;

  auto good = [&](const vector<int>& p) {
    set<int> d1, d2;
    for (int r = 0; r < (int)(8); ++r) {
      if (R[r] != -1 && R[r] != p[r]) return false;
      d1.insert(p[r] + r);
      d2.insert(p[r] - r);
    }
    return d1.size() == 8 && d2.size() == 8;
  };

  do {
    if (good(perm)) {
      for (int i = 0; i < (int)(8); ++i) {
        for (int j = 0; j < (int)(8); ++j) cout << (perm[i] == j ? 'Q' : '.');
        cout << endl;
      }
      break;
    }
  } while (next_permutation((perm).begin(), (perm).end()));

  return 0;
}

