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>

using namespace std;

#define Free -1
#define N_free 1
#define N 8

int row[N],col[N],dpos[2*N-1],dneg[2*N-1];
bool p[N][N];
void clear(){
  for(int i=0;i<N;i++){
    row[i]=Free;
    col[i]=Free;
  }
  for(int i=0;i<2*N-1;i++){
    dpos[i]=dneg[i]=Free;
}
}

void print(){
  for(int i=0;i< N;i++){
    for(int j=0; j<N;j++){
      if(p[i][j]== true){
	if(row[i]!= j){
	  return;
	}
      }
    }
  }
    for(int i=0;i<N;i++){
      for(int j=0;j<N;j++){
	if(row[i]== j){
	  cout<<"Q";
	}
	else cout<<".";
      }
      cout<<endl;
    }
  }

  void recursive(int i){

    if(i==N){
      print();
      return ;
    }

    for(int j=0;j<N ;j++){
      if(col[j]== N_free || dpos[i+j]== N_free|| dneg[i-j+N-1]==N_free){
	continue;
      }
      row[i]=j;
      col[j]=dpos[i+j]=dneg[i-j+N-1]=N_free;
      recursive(i+1);
      row[i]=col[j]=dpos[i+j]=dneg[i-j+N-1]= Free;
    }
  }

int main(){
  clear();
  for(int i=0;i<N;i++){
    for(int j=0;j<N;j++){
      p[i][j]=false;
    }
  }
  int n=0;
  cin>> n;
  for(int i=0; i< n;i++){
    int r,c;
    cin>> r>> c;
    p[r][c]=true;
  }
  recursive(0);

  return 0;
}

