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<iostream>
#define n 8
using namespace std;
int row[n],col[n],dpos[2*n-1],dneg[2*n-1],X[n][n];
void print()
{
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
        if(X[i][j]&&row[i]!=j) return ; //这一段留着还是必要的，因为每到8 会输出，不判断就很难受
    for(int i=0;i<n;cout<<endl,i++)
        for(int j=0;j<n;j++)
        cout<<((row[i]==j)?"Q":".");
}
void recursive(int i)
{
    if(i==n)
    {
        print();
        return ;
    }
    for(int j=0;j<n;j++)
    {
        if(col[j]==1||dpos[i+j]==1||dneg[i-j+n-1]==1)continue;
        row[i]=j; col[j]=dpos[i+j]=dneg[i-j+n-1]=1;
        recursive(i+1);
        row[i]=col[j]=dpos[i+j]=dneg[i-j+n-1]=-1;
    }
}
int main()
{
    for(int i=0;i<n;i++)row[i]=col[i]=-1;
    for(int i=0;i<2*n-1;i++)dpos[i]=dneg[i]=-1;
    int k,x,y;
    cin>>k;
    for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)X[i][j]=0;
    for(int i=0;i<k;i++)cin>>x>>y,X[x][y]=1;
    recursive(0);
    return 0;
}

