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;

int main() {
    int k;
    cin >> k;
    vector<int> fix(8, -1);
    for (int i = 0; i < k; ++i) {
        int r, c;
        cin >> r >> c;
        fix[r] = c;
    }
    vector<int> ord(8);
    iota(ord.begin(), ord.end(), 0);
    do {
        bool ng = false;
        for (int i = 0; i < 8; ++i) {
            if (fix[i] >= 0 && fix[i] != ord[i]) ng = true;
        }

        if (ng) continue;
        for (int i = 0; i < 8; ++i) {
            for (int j = i + 1; j < 8; ++j) {
                if (j - i == ord[j] - ord[i]) ng = true;
                if (j - i == ord[i] - ord[j]) ng = true;
            }
        }
        if (ng) continue;
        for (int i = 0; i < 8; ++i) {
            for (int j = 0; j < 8; ++j) {
                if (j == ord[i]) cout << "Q";
                else cout << ".";
            }
            cout << endl;
        }
    } while (next_permutation(ord.begin(), ord.end()));
}
