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...
import itertools


def hand_in_answer(v):
    base = [".", ".", ".", ".", ".", ".", ".", "."]
    for i in range(8):
        tmp = base.copy()
        tmp[v[i]] = "Q"
        print("".join(tmp))


def ifGoodPosition(position, v):
    yo = position[0]
    xo = position[1]

    x = xo
    y = yo
    x += 1
    y += 1
    while x < 8 and x >= 0 and y >= 0 and y < 8:
        if v[y] == x:
            return False
        x += 1
        y += 1

    x = xo
    y = yo
    x += 1
    y -= 1
    while x < 8 and x >= 0 and y >= 0 and y < 8:
        if v[y] == x:
            return False
        x += 1
        y -= 1

    x = xo
    y = yo
    x -= 1
    y += 1
    while x < 8 and x >= 0 and y >= 0 and y < 8:
        if v[y] == x:
            return False
        x -= 1
        y += 1

    x = xo
    y = yo
    x -= 1
    y -= 1
    while x < 8 and x >= 0 and y >= 0 and y < 8:
        if v[y] == x:
            return False
        x -= 1
        y -= 1

    return True


fixed_q_count = int(input())
x_choices = set([_ for _ in range(8)])
fixed_qs = []
for _ in range(fixed_q_count):
    y, x = map(int, input().split(" "))
    fixed_qs.append([y, x])

for v in itertools.permutations(x_choices, 8):
    flag2 = True
    for fixed_q in fixed_qs:
        if not v[fixed_q[0]] == fixed_q[1]:
            flag2 = False
            break

    if not flag2:
        continue

    flag = True
    for y in range(8):
        if not ifGoodPosition([y, v[y]], v):
            flag = False
            break

    if flag:
        hand_in_answer(v)
        break

