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 int long long
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

signed main() {
    int n;
    cin >> n;

    map<int,int> mp;
    rep(i,8) mp[i] = -1;
    rep(i,n){
        int x,y;
        cin >> x >> y;
        mp[x] = y;
    }

    vector<int> v(8);
    rep(i,8) v[i] = i;

    vector<vector<char> > ans(8,vector<char>(8,'.'));
    do{
        bool flag = true;
        for(int i = 0; i < 8; i++){
            if(mp[i] == -1) continue;
            else {
                if(mp[i] != v[i]){
                    flag = false;
                    break;
                }
            }
        }

        for(int i = 0; i < 8; i++){
            for(int j = i + 1; j < 8; j++){
                if(j - i == abs(v[j] - v[i])) {
                    flag = false;
                    break;
                }
            }
        }

        if(flag) {
            for(int i = 0; i < 8; i++){
                ans[i][v[i]] = 'Q';
            }
            break;
        }
    } while(next_permutation(v.begin(),v.end()));

    rep(i,8) {
        rep(j,8){
            cout << ans[i][j];
        }
        cout << endl;
    }
}

