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 N_FREE 1

int Row[N], Col[N], dpos[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++)
  {
    dpos[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(N_FREE == Col[j] ||
       N_FREE == dpos[i+j] ||
       N_FREE == dneg[i-j+N-1])
      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(){
  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;
}

