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 <cstdio>
#define N 8
#define FREE -1
#define NOT_FREE 1

using namespace std;

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

void print () {
  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 recursive (int i) {
  // printf("i = %d, row[%d] = %d\n", i, i, row[i]);
  if (i == N) {
    print();
    return;
  }

  if (row[i] != FREE) {
    // cout << "next" << endl;
    recursive(i + 1);
    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;
    recursive(i + 1);
    // 回溯
    row[i] = col[j] = dpos[i + j] = dneg[i - j + (N - 1)] = FREE;
  }
}

int main () {
  int k, _i, _j;

  cin >> k;

  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 < k; i++) {
    cin >> _i >> _j;
    row[_i] = _j;
    col[_j] = dpos[_i + _j] = dneg[_i - _j + (N - 1)] = NOT_FREE;
  }

  recursive(0);

  return 0;
}
