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;

const int N=8,FREE=-1,NOT_FREE=1;
int row[8],col[8],dpos[15],dneg[15];
bool X[8][8];

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

        row[i]=j; col[j]=dpos[i+j]=dneg[i-j+N-1]=NOT_FREE;
        recursive(i+1);
        row[i]=col[j]=dpos[i+j]=dneg[i-j+N-1]=FREE;
    }
}

int main(){

    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);
}
