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 row[8];
bool X[8][8];
int n;


void print()
{
        for ( int i = 0; i < 8; i++ ) {
                for ( int j = 0; j < 8; j++ ) {
                        if ( X[i][j] ) {
                                if ( row[i] != j ) return;
                        }
                }
        }
        
        for ( int i = 0; i < 8; i++ ) {
                for ( int j = 0; j < 8; j++ ) {
                        cout << (row[i] == j ? 'Q' : '.');
                }
                cout << endl;
        }
}
void dfs(int i, int left = 0, int center = 0, int right = 0)
{
        if ( i == 8 ) {
                print();
                return;
        }
        
        for ( int j = 0; j < 8; j++ ) {
                if ( ((left >> j) & 1) || ((center >> j) & 1) || ((right >> j) & 1) ) continue;
                row[i] = j;
                dfs(i + 1, (left | (1 << j)) << 1, center | (1 << j), (right | (1 << j)) >> 1);
                row[i] = -1;
        }
}

int main()
{

        cin >> n;
        for ( int i = 0; i < n; i++ ) {
                int r, c;
                cin >> r >> c;
                X[r][c] = true;
        }

        dfs(0);

        return ( 0 );
}

