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
tate = [True]*N
yoko = [True]*N
rnaname = [True]*(2*N-1)
lnaname = [True]*(2*N-1)
board = [[True]*N for i in range(N)]


def update(x, y, value):
    tate[x] = value
    yoko[y] = value
    rnaname[x+y] = value
    lnaname[x-y+N-1] = value
    board[x][y] = value


def check(x, y):
    if tate[x] and yoko[y] and rnaname[x+y] and lnaname[x-y+N-1]:
        return True
    return False


def show():
    for i in range(N):
        for j in range(N):
            print('.' if board[i][j] else 'Q', sep='', end='')
        print()


# iは縦ベース
def solve(i):
    if i == N:
        show()
        exit(0)
    if tate[i] is False:
        solve(i+1)
    for j in range(N):
        if check(i, j):
            update(i, j, False)
            solve(i+1)
            update(i, j, True)


n = int(input())
for i in range(n):
    Q = list(map(int, input().split()))
    update(Q[0], Q[1], False)

for i in range(N):
    if tate[i]:
        solve(i)
        break
solve(N)


