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 FREE -1
#define NOT_FREE 1

void A();
void print();
void queen(int);

int X[8],Y[8],dpos[15],dneg[15];

bool Q[8][8];

int main(){
  A();

  for(int i=0;i<8;i++){
    for(int j=0;j<8;j++){
      Q[i][j]=false;
    }
  }

  int k;
  cin >> k;

  for(int i=0;i<k;i++){
    int r,c;
    cin >> r >> c;
    Q[r][c]=true;
  }
  queen(0);

  return 0;
}

void A(){
  for(int i=0;i<8;i++){
    X[i]=FREE;
    Y[i]=FREE;
  }
  for(int i=0;i<15;i++){
    dpos[i]=FREE;
    dneg[i]=FREE;
  }
}

void print(){
  for(int i=0;i<8;i++){
    for(int j=0;j<8;j++){
      if(Q[i][j]){
	if(X[i]!=j)return;
      }
    }
  }

  for(int i=0;i<8;i++){
    for(int j=0;j<8;j++){
      cout << ((X[i]==j) ? "Q" : ".");
    }
    cout << endl;
  }
}

void queen(int a){
  
  if(a == 8){
    print();
    return;
  }

  for(int i=0;i<8;i++){
    if(NOT_FREE == Y[i] || NOT_FREE == dpos[a+i] || NOT_FREE == dneg[a-i+7])continue;

    X[a]=i;
    Y[i]=dpos[a+i]=dneg[a-i+7]=NOT_FREE;

    queen(a+1);

    X[a]=Y[i]=dpos[a+i]=dneg[a-i+7]=FREE;
  }
}

