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 as per
z=int(input())
chk=[]
for i in range(z):
    chk.append([int(i) for i in input().split()])
a=list(range(8))
pera=list(per(a))
k=0
for i in pera:
    st1=set() ; st2=set() ; flag=0
    for j in range(z):
        if i[chk[j][0]]!=chk[j][1]:
            flag=1
            break
    if flag==1: continue
    for j in range(8):
        if i[j]-j not in st1:
            st1.add(i[j]-j)
        else: break
        if i[j]+j not in st2:
            st2.add(i[j]+j)
        else: 
            break
    if len(st1)==8 and len(st2)==8:
        ans=[['.']*8 for i in range(8)]
        for j in range(8):
            ans[j][i[j]]='Q'
        for le in range(8):
            print(''.join(ans[le]))
        break
                
        
