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


def putQueen(i):
    if i == N:
        printBoard()
        return

    if i in default:
        putQueen(i+1)

    else:
        for j in range(N):
            if col [j]   or \
               dpos[i+j] or \
               dneg[i-j+N-1]:
                continue

            tmp = row[i]
            row[i] = j
            col[j] = dpos[i+j] = dneg[i-j+N-1] = True

            putQueen(i+1)

            row[i] = tmp
            col[j] = dpos[i+j] = dneg[i-j+N-1] = False


def printBoard():
    for i in range(N):

        line = ''
        for j in range(N):
            if row[i] == j: line += 'Q'
            else          : line += '.'

        print(line)


if(__name__ == '__main__'):
    row  = [None] * N
    col  = [False] * N
    dpos = [False] * (2*N - 1)
    dneg = [False] * (2*N - 1)

    default = []

    n = int(input())

    for _ in range(n):
        r, c = [int(i) for i in input().split()]

        row[r] = c
        col[c] = dpos[r+c] = dneg[r-c+N-1] = True

        default.append(r)

    putQueen(0)

