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.

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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...
FREE = -1
NOT_FREE = 1

N = 8
B = [["."] * N for _ in range(N)]
row = [FREE] * N
col = [FREE] * N
dpos = [FREE] * (2 * N - 1)
dneg = [FREE] * (2 * N - 1)

def update(r, c, FREE_OR_NOT):
    B[r][c] = "." if FREE_OR_NOT == FREE else "Q"
    row[r] = col[c] = dpos[r + c] = dneg[r - c + N - 1] = FREE_OR_NOT

def draw():
    for i in range(N):
        print("".join(B[i]))

def recursive(i):
    if i == N:
        draw()
        return
    if "Q" in B[i]:
        recursive(i + 1)

    for j in range(N):
        if NOT_FREE in [row[i], col[j], dpos[i + j], dneg[i - j + N - 1]]:
            continue
        else:
            update(i, j, NOT_FREE)
        recursive(i + 1)
        update(i, j, FREE)

k = int(input())

for _ in range(k):
    r, c = map(int, input().split())
    update(r, c, NOT_FREE)
recursive(0)
