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 static int N = 8;
const static int FREE = -1;
const static int NOT_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(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);


  return 0;
}

