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 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] = col[i] = FREE;
  for(int i = 0;i < 2*N-1;i++) dpos[i] = 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 recusive(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;
      recusive(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;
    }

    recusive(0);

    return 0;
}

