Problem:
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8 Queens Problem

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.

For a given chess board where $k$ queens are already placed, find the solution of the 8 queens problem.

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 '.'.

Constraints

There is exactly one solution

Sample Input 1

2
2 2
5 3

Sample Output 1

......Q.
Q.......
..Q.....
.......Q
.....Q..
...Q....
.Q......
....Q...

Solution in Scala (judge verdict: Accepted)

import scala.io.StdIn

object Main extends App {

  val matrixSize = 8
  val free = -1
  val notFree = 1

  val rows = Array.fill(matrixSize)(free)
  val cols = Array.fill(matrixSize)(free)
  val dpos = Array.fill(2 * matrixSize - 1)(free)
  val dneg = Array.fill(2 * matrixSize - 1)(free)

  val x = Array.fill(8, 8)(false)


  val n = StdIn.readLine().toInt
  (0 until n).foreach { _ =>
    val Array(r, c) = StdIn.readLine().split(" ").map(_.toInt)
    x(r)(c) = true
  }

  def printBoard(): Unit = {
    val rc = for (r <- 0 until matrixSize;
                  c <- 0 until matrixSize if x(r)(c)
    ) yield (r, c)
    val judge = rc.forall(e => rows(e._1) == e._2)
    if (judge) {
      for (r <- 0 until matrixSize;
           c <- 0 until matrixSize) {
        print(if (rows(r) == c) "Q" else ".")
        if (c == matrixSize - 1) println()
      }
    }
  }

  def recursive(i: Int): Unit = {
    if (i == matrixSize) {
      printBoard()
    } else {
      for (j <- 0 until matrixSize) {
        if (cols(j) != notFree
          && dpos(i + j) != notFree
          && dneg(i - j + matrixSize - 1) != notFree) {
          rows(i) = j
          cols(j) = notFree
          dpos(i + j) = notFree
          dneg(i - j + matrixSize - 1) = notFree
          recursive(i + 1)
          rows(i) = free
          cols(j) = free
          dpos(i + j) = free
          dneg(i - j + matrixSize - 1) = free

        }
      }
    }
  }

  // main start
  recursive(0)

}