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>
#include <cassert>
using namespace std;
#define N (8)
#define FREE (-1)
#define NOT_FREE (1)

int row[N],col[N],drops[2*N-1],dneg[2*N-1];

bool X[N][N];

void initialize(){
    for(int i=0;i<N;i++){
        row[i]=FREE;col[i]=FREE;
    }
    for(int i=0;i<2*N-1;i++){
        drops[i]=FREE;dneg[i]=FREE;
    }
}

void printBoard(){
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            if(X[i][j]){
                if(row[i]!=j)return;
            }
        }
    }
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            cout<<((row[i]==j)?"Q":".");
        }
        cout<<endl;
    }
}

void recursive(int i){
    if(i==N){
        printBoard();return;
    }
    
    for(int j=0;j<N;j++){
        if(NOT_FREE==col[j]||NOT_FREE==drops[i+j]||NOT_FREE==dneg[i-j+N-1])continue;
        row[i]=j;col[j]=drops[i+j]=dneg[i-j+N-1]=NOT_FREE;
        recursive(i+1);
        row[i]=col[j]=drops[i+j]=dneg[i-j+N-1]=FREE;
        
    }
}

int main(void){
    initialize();
    for(int i=0;i<N;i++){
        for(int j=0;j<N;j++){
            X[i][j]=false;
        }
    }
    
    int k;cin>>k;
    for(int i=0;i<k;i++){
        int r,c;cin>>r>>c;
        X[r][c]=true;
    }
    recursive(0);
    return 0;
}

