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 N 8
#define FREE 1
#define NOT_FREE -1

bool Q[N][N];
int row[N];
int col[N];
int dpos[2*N-1];
int dneg[2*N-1];


void putQueen(int);
void printBoard(void);

void printBoard(){
  for(int i = 0; i < N; i++){
    for(int j = 0; j < N; j++){
      if(Q[i][j] && 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 putQueen(int i){
  if(i == N){
    printBoard();
    return;
  }
  for(int j = 0; j < N; j++){
    if(col[j] == NOT_FREE || dpos[i+j] == NOT_FREE || dneg[i-j+N-1] == NOT_FREE){
      continue;
    }
    row[i] = j;
    col[j] = dpos[i+j] = dneg[i-j+N-1] = NOT_FREE;
    putQueen(i+1);
    col[j] = dpos[i+j] = dneg[i-j+N-1] = FREE;
  }
}
signed main(){
  int k,r,c;

  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;
  }

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

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

  putQueen(0);
}

