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_ = 8
FREE_ = -1
NOTFREE_ = 1
row = [FREE_ for i in range(N_)]
col = [FREE_ for i in range(N_)]
dpos = [FREE_ for i in range(2*N_-1)]
dneg = [FREE_ for i in range(2*N_-1)]
X = [[False for i in range(N_)] for j in range(N_)]

def output():
    for i in range(N_):
        for j in range(N_):
            if X[i][j]:
                if row[i] != j:
                    return

    for i in range(N_):
        for j in range(N_):
            if row[i] == j:
                print('Q', end='')
            else:
                print('.', end='')
        print()

def recursive(i):
    if i == N_:
        output()
        return

    for j in range(N_):
        if col[j] == NOTFREE_ or dpos[i + j] == NOTFREE_ or dneg[i - j + N_ - 1] == NOTFREE_:
            continue
        row[i] = j
        col[j] = dpos[i + j] = dneg[i - j + N_ - 1] = NOTFREE_
        recursive(i + 1)
        row[i] = col[j] = dpos[i + j] = dneg[i - j + N_ - 1] = FREE_

k = int(input())
for i in range(k):
    r,c = map(int, input().split())
    X[r][c] = True

recursive(0)

