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 itertools import permutations

k = int(input())
r = [0] * k; c = [0] * k;
for i in range(k):
    r[i], c[i] = [int(a) for a in input().split()]

r_avail = list(range(8))
c_avail = list(range(8))

for i in range(k):
    r_avail.remove(r[i])
    c_avail.remove(c[i])

cperms = list(permutations(c_avail, 8 - k))

for perm in cperms:
    Q = [(r[i], c[i]) for i in range(k)] + list(zip(r_avail, perm))
    for a in range(8):
        for b in range(a + 1, 8):
            rdiff = Q[a][0] - Q[b][0]
            cdiff = Q[a][1] - Q[b][1]
            if abs(rdiff) == abs(cdiff):
                break # 斜めの線上にある
        else:
            continue
        break
    else: # 斜めの線上にあるQueenは無い
        break

ans = [['.' for _ in range(8)] for _ in range(8)]
for i in range(8):
    ans[Q[i][0]][Q[i][1]] = 'Q'

for i in range(8):
    print("".join(ans[i]))
