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...
import itertools
queen=[-1]*8
N=int(input())
de=[]
for i in range(N):
    x,y =list(map(int,input().split()))
    de.append(x)
    queen[x]=y
def check(n):#nはクイーンの個数
    for y in range(n):
        if crshu(queen[y],y):return False
    return True
def crshu(x,y):
    for y1 in range(y):
        x1=queen[y1]
        if x1-y1==x-y or x1+y1==x+y:return True
    return False
def board():
    king=[]
    for i in range(8):
        king.append([i,queen[i]])
    for i in range(8):
        for j in range(8):
            if [i,j] in king:
                print("Q",end="")
            else:
                print(".",end="")
        print()
    exit()

def search(n,y):
    if n==y:
        if check(8):
            board()
    else:
        if y in de:
            search(n,y+1)
        else:
            for i in range(n):
                if i not in queen :
                    queen[y]=i
                    search(n,y+1)
                    for l in range(y+1,8):
                        if l not in de:
                            queen[l]=-1
search(8,0)

