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...
def queenprint() :
    global x
    global col
    global row
    global k
    global maxx
    
    for i in range(maxx):
        for j in range(maxx):
            if x[i][j] and row[i] != j:
                return
       
    for i in range(maxx):
        for j in range(maxx):
            if row[i]==j: 
                print('Q',end='')
            else:
                print('.',end='')
        print()
        
def recursive(i):
    global x
    global col
    global row
    global k
    global maxx
    
    if (i == maxx) :
        queenprint()
        return
    
    for j in range(maxx):
        # 不适合放在这里
        if (col[j] or dpos[i + j] or dneg[i - j + maxx - 1]):
            continue
        # 尝试放下这里
        row[i] = j 
        col[j] = dpos[i + j] = dneg[i - j + maxx - 1] = 1;
        recursive(i + 1);
        # 恢复刚才的操作继续其他的操作
        row[i] = col[j] = dpos[i + j] = dneg[i - j + maxx - 1] = 0

maxx = 8    
k=int(input())
x=[[ 0 for i in range(maxx)] for j in range(maxx)]
col=[0 for i in range(maxx)]
row=[0 for i in range(maxx)]
dpos=[0 for i in range(maxx * 2 - 1)]
dneg=[0 for i in range(maxx * 2 - 1)]
      
for i in range(k):
    r,c=[int(x) for x in input().split()]
    x[r][c] = 1
   
recursive(0)   
