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 <iostream>
#include <vector>
#include <algorithm>
#define rep(i, n) for(int i = 0; i < (int)(n); ++i)
#define REP(i, n) for(int i = 1; i < (int)(n); ++i)
using namespace std;

int main() {
    int k;
    cin >> k;
    vector<int> a(k), b(k);
    rep(i, k) cin >> a[i] >> b[i];
    
    vector<int> vec = {0, 1, 2, 3, 4, 5, 6, 7};
    do {
        bool judge = true;
        rep(i, k) {
            if (vec[a[i]] != b[i]) judge = false;
        }
        if (!judge) continue;
        rep(i, 8) rep(j, 8) {
            if (i == j) continue;
            if (abs(i - j) == abs(vec[i] - vec[j])) judge = false;
        }
        if (judge) {
            vector<vector<char>> ans(8, vector<char>(8, '.'));
            rep(i, 8) ans[i][vec[i]] = 'Q';
            rep(i, 8) {
                rep(j, 8) cout << ans[i][j];
                cout << endl;
            }
            return 0;
        }
    } while (next_permutation(vec.begin(), vec.end()));
}

