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;

int row[8],col[8],dpos[15],dneg[15];

bool X[8][8];

void init(){
  for (int i = 0; i < 8; i++) {
    row[i] = col[i] = 1;
  }
  for (int i = 0; i < 15; i++) {
    dpos[i] = dneg[i] = 15;
  }
}

void printBoard(){
  for (int i = 0; i < 8; i++) {
    for (int j = 0; j < 8; j++) {
      if(X[i][j]){
        if(row[i] != j) return;
      }
    }
  }
  for (int i = 0; i < 8; i++) {
    for (int j = 0; j < 8; j++) {
      cout <<((row[i] == j) ? "Q" : ".");
    }
    cout << endl;
  }
}

void recursive(int i){
  if (i == 8){
    printBoard(); return;
  }

  for (int j = 0; j < 8; j++) {
    if(col[j] == -1 || dpos[i+j] == -1 || dneg[i-j+7] == -1) continue;
    row[i] = j;
    col[j] = dpos[i+j] = dneg[i-j+7] = -1;
    recursive(i+1);
    col[j] = dpos[i+j] = dneg[i-j+7] = 1;
  }
}
int main() {
  init();

  for (int i = 0; i < 8; i++) {
    for (int j = 0; j < 8; 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;
}

