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 mat[8];

void print()
{
    for(int i=0;i<8;i++)
    {
        for(int j=0;j<8;j++)
        {
            if(mat[i]==j)
            {
                cout<<'Q';
            }else
            {
                cout<<'.';
            }
        }
        cout<<endl;
    }
}
bool dfs(int r)
{
    if(r==8)
    {
        print();
        return true;
    }

    if(mat[r]!=-1)
    {
        for(int i=0;i<r;i++)
        {
            if(mat[i]==mat[r]||r-i==mat[r]-mat[i]||r-i==mat[i]-mat[r])
            {
                return false;
            }
        }
        return dfs(r+1);
    }
    for(int i=0;i<8;i++)
    {
        mat[r]=i;
        bool flag=true;
        for(int j=0;j<r;j++)
        {
            if(mat[j]==i||r-j==i-mat[j]||r-j==mat[j]-i)
            {
                flag=false;
                break;
            }
        }
        if(flag&&dfs(r+1))
        {
            return true;
        }
    }
    mat[r]=-1;
    return false;
}

int main()
{
    int n;

    int r,c;

    ios::sync_with_stdio(false);

    cin>>n;
    memset(mat,-1,sizeof(mat));
    for(int i=0;i<n;i++)
    {
        cin>>r>>c;
        mat[r]=c;
    }

    dfs(0);

    return 0;
}

