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...
from sys import stdin
N = 8
FREE = -1
NOT_FREE = 1
row = [FREE] * N
col = [FREE] * N
dpos = [FREE] * (2 * N - 1)
dneg = [FREE] * (2 * N - 1)
X = [[False] * N for _ in range(N)]
n = int(stdin.readline())
for _ in range(0, n):
    r, c = map(int, stdin.readline().split())
    X[r][c] = True
def printBoard():
    for i in range(0, N):
        for j in range(0, N):
            if X[i][j] != 0 and row[i] != j: return
    for i in range(0, N):
        print("." * row[i] + "Q" + "." * (8 - row[i] - 1))
def recursive(i):
    if i == N:
        printBoard()
        return
    for j in range(0, N):
        if (NOT_FREE == col[j] or
            NOT_FREE == dpos[i + j] or
            NOT_FREE == dneg[i - j + N - 1]): continue
        row[i] = j
        col[j] = dpos[i + j] = dneg[i - j + N - 1] = NOT_FREE
        recursive(i + 1)
        row[i] = col[j] = dpos[i + j] = dneg[i - j + N - 1] = FREE
recursive(0)