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 <string>
#include <stack>
using namespace std;

int a[8][8];
int row[8]={};
int col[8]={};
int ru[15]={};
int rd[15]={};

void set(int r,int c,int n){
  a[r][c]=n;
  row[r]=n;
  col[c]=n;
  ru[r+c]=n;
  rd[7+c-r]=n;
}

void unset(int r,int c){
  a[r][c]=0;
  row[r]=0;
  col[c]=0;
  ru[r+c]=0;
  rd[7+c-r]=0;
}

bool can_set(int r,int c){
  return !(row[r]>0 ||
           col[c]>0 ||
           ru[r+c]>0 ||
           rd[7+c-r]>0);
}

void print(){
  for(int r=0;r<8;r++){
    for(int c=0;c<8;c++){
      if(a[r][c]>0){
        cout<<'Q';
      }else{
        cout<<'.';
      }
    }
    cout<<endl;
  }
}

void solve(int r){
  if(r==8){
    print();
    return;
  }
  if(row[r]>0) solve(r+1);
  for(int c=0;c<8;c++){
    if(can_set(r,c)){
      set(r,c,1);
      solve(r+1);
      unset(r,c);
    }
  }
}

int main(){
  int k;cin>>k;

  for(int i=0;i<8;i++)
    for(int j=0;j<8;j++) a[i][j]=0;

  for(int i=0;i<k;i++){
    int r,c; cin>>r>>c;
    set(r,c,2);
  }

  solve(0);
}