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...
class Field:

    def __init__(self, N):
        self.N = N
        self.row = [-1 for _ in range(N)]
        self.hold = []
        self.col = [0 for _ in range(N)]
        self.dpos = [0 for _ in range(2*N - 1)]
        self.dneg = [0 for _ in range(2*N - 1)]

    def putQueen(self, i):
        if i in self.hold:
            self.putQueen(i + 1)
            return 
        if i == self.N:
            self.printBoard()
            return
        for j in range(self.N):
            if self.col[j] or self.dpos[i+j] or self.dneg[i-j+self.N-1]:
                continue
            self.setQueen(i, j)
            self.putQueen(i + 1)
            self.col[j] = self.dpos[i+j] = self.dneg[i-j+self.N-1] = 0

    def setQueen(self, i, j):
        self.row[i] = j
        self.col[j] = self.dpos[i + j] = self.dneg[i - j + self.N - 1] = 1

    def printBoard(self):
        print('\n'.join("." * r + "Q" + "." * (self.N - r - 1) for r in self.row))


if __name__ == "__main__":
    field = Field(8)
    for _ in range(int(input())):
        i, j = (map(int, input().split()))
        field.setQueen(i, j)
        field.hold.append(i)
    field.putQueen(0)
