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
FREE = -1
NOT_FREE = 1

row = [FREE for _ in range(N)]
col = [FREE for _ in range(N)]
dpos = [FREE for _ in range(2*N-1)]
dneg = [FREE for _ in range(2*N-1)]

X = [[False for _ in range(N)] for _ in range(N)]

def printBoard():
    i,j = 0, 0
    for i in range(N):
        for j in range(N):
            if X[i][j]:
                if row[i] != j: return
    
    for i in range(N):
        for j in range(N):
            print("Q" if row[i]==j else ".", end='')
        print()
            

def recursive(i):
    j = 0

    if i==N:
        printBoard()
    
    else:
        for j in range(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

k = int(input())
for _ in range(k):
    r,c = map(int,input().split())
    X[r][c] = True

recursive(0)

