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...
row, col = [True] * 8, [True] * 8
d_pos, d_neg = [True] * 15, [True] * 15
board = [["."] * 8 for _ in range(8)]


def change_state(r: int, c: int, undo: bool = False) -> None:
    if undo:
        row[r] = col[c] = d_pos[r + c] = d_neg[r - c + 7] = True
        board[r][c] = "."
    else:
        row[r] = col[c] = d_pos[r + c] = d_neg[r - c + 7] = False
        board[r][c] = "Q"


def solve_eight_queens(r: int = 0) -> None:
    while r < 8 and not row[r]:
        r += 1
    if r == 8:
        for board_row in board:
            print(*board_row, sep="")
        return
    for c in range(8):
        if not col[c] or not d_pos[r + c] or not d_neg[r - c + 7]:
            continue
        change_state(r, c)
        solve_eight_queens(r + 1)
        change_state(r, c, undo=True)


def main():
    K, *RC = map(int, open(0).read().split())
    for r, c in zip(*[iter(RC)] * 2):
        row[r] = col[c] = d_pos[r + c] = d_neg[r - c + 7] = False
        board[r][c] = "Q"
    solve_eight_queens()


if __name__ == "__main__":
    main()

