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 <math.h>
#include <iostream>
#include<string>
#include <algorithm>
#include <bits/stdc++.h>
#include <vector>
#include <stdlib.h>
using namespace std;

int main()
{
int k;cin >> k;bool good=false;bool check=true;
vector<int> r(k);
vector<int> c(k);
for(int i=0;i<k;i++)cin >> r[i] >> c[i];
vector<int> row(8);vector<int> column(8);
for(int i=0;i<8;i++){row[i]=i;column[i]=i;}
do{
check=true;good=true;
for(int i=0;i<k;i++){if(!check)break;check=false;for(int j=0;j<8;j++)if(r[i]==row[j]&&c[i]==column[j]){check=true;break;}}
if(!check)continue;
for(int i=0;i<8;i++)for(int j=i+1;j<8;j++)if(abs(row[i]-row[j])==abs(column[i]-column[j])){good=false;break;}
if(good)break;
}while(next_permutation(column.begin(),column.end()));
vector<vector<char>> board(8,vector<char>(8,'.'));
for(int i=0;i<8;i++)board[row[i]][column[i]]='Q';
for(int i=0;i<8;i++){for(int j=0;j<8;j++)cout << board[i][j];cout << endl;}
}

