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;
#define rep(i, n) for(int i = 0, i##_len = (n); i < i##_len; ++i)
#define rep2(i, x, n) for(int i = x, i##_len = (n); i < i##_len; ++i)
#define all(n) begin(n), end(n)
using ll = long long;
using P = pair<int, int>;
using vi = vector<int>;
using vl = vector<ll>;
using vs = vector<string>;
using vc = vector<char>;
using vb = vector<bool>;
vi dir = {-1, 0, 1, 0, -1, -1, 1, 1, -1};

int main() {
  int k;
  cin >> k;
  vector<P> p;
  rep(i, k) {
    int r, c;
    cin >> r >> c;
    p.push_back(P(r, c));
  }
  vi v = {0, 1, 2, 3, 4, 5, 6, 7};
  do {
    bool ok = true;
    rep(i, k) {
      int x = p[i].first;
      int y = p[i].second;
      if(v[x] != y) ok = false;
    }
    rep(i, 7) {
      int x = i, y = v[i];
      rep2(j, i + 1, 8) {
        int nx = j, ny = v[j];
        if(abs(x - nx) == abs(y - ny)) ok = false;
      }
    }
    if(ok) break;
  } while(next_permutation(all(v)));
  rep(i, 8) {
    rep(j, 8) cout << (v[i] == j ? 'Q' : '.');
    cout << endl;
  }
}
