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<cstdio>
#include<cstring>
#include<vector>
using namespace std;
const int N=8;
int row[N],col[N],lt[N*N],rb[N*N];
int n;
int ans[N][N];
void print()
{
    for(int r=0; r<N; r++)
    {
        for(int c=0; c<N; c++)
        {
            if(ans[r][c]==1)
            {
                if(row[r]!=c) return;
            }
        }
    }
    for(int r=0; r<N; r++)
    {
        for(int c=0; c<N; c++)
        {

            if(row[r]==c) printf("Q");
            else printf(".");

        }
        printf("\n");
    }
}
void recursive(int r)
{
    if(r==N)
    {
        print();
        return;
    }
    for(int c=0; c<N; c++)
    {
        if( col[c]==1||lt[r+c]==1||rb[c-r]==1) continue;
        row[r]=c;
        col[c]=lt[r+c]=rb[c-r]=1;
        recursive(r+1);
        row[r]=col[c]=lt[r+c]=rb[c-r]=-1;
    }
}

int main(void)
{
//    freopen("in.txt","r",stdin);
    scanf("%d",&n);
    int r,c;
    for(int i=0; i<N; i++)
    {
        row[i]=col[i]=-1;
    }
    for(int i=0; i<2*N; i++)
    {


        lt[i]=rb[i]=-1;

    }
    while(n--)
    {
        scanf("%d%d",&r,&c);
        ans[r][c]=1;

    }
    recursive(0);


    return 0;
}