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;

using ll=long long;
const ll INF=1LL<<60;
const ll MOD=1e9+7;

int main(){
    vector<vector<char>> board(8, vector<char>(8));
    vector<vector<bool>> check(8, vector<bool>(8, true));
    int k; cin >> k;
    vector<int> r(k), c(k);
    for(int i=0; i<k; i++){
        cin >> r[i] >> c[i];
    }
    vector<int> order(8);
    for(int i=0; i<8; i++) order[i]=i;
    do{
        bool ok=true;
        for(int i=0; i<k; i++){
            if(order[c[i]] != r[i]) ok=false;
        }
        if(!ok) continue;
        for(int i=0; i<7; i++){
            for(int j=1; i+j<8; j++){
                if(order[i+j] == order[i]+j || order[i+j] == order[i]-j) ok=false;
            }
        }
        if(ok) break;
    }while(next_permutation(order.begin(), order.end()));
    for(int i=0; i<8; i++){
        for(int j=0; j<8; j++){
            if(order[j] == i) cout << "Q";
            else cout << ".";
        }
        cout << endl;
    }
    return 0;
}


