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...
n = int(input())
nt = []

board = [["." for _ in range(8)] for _ in range(8)]

for i in range(n):
    y, x = map(int, input().split())
    board[y][x] = "Q"
    nt.append(y)

def output():
    for row in board:
        print("".join(row), sep=None)

def check(x, y):
    for i in range(8):
        for j in range(8):
            if i == y and j == x:
                continue
            elif i == y and board[i][j] == "Q":
                return False
            elif j == x and board[i][j] == "Q":
                return False
            elif abs(i-y) == abs(j-x) and board[i][j] == "Q":
                return False
    return True

def dfs(row):
    if row == 8:
        output()
        return
    if row in nt:
        dfs(row+1)
        return

    for i in range(8):
        board[row][i] = "Q"
        if not check(i, row):
            board[row][i] = "."
            continue
        else:
            dfs(row+1)
            board[row][i] = "."


dfs(0)