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...
#include <cstdio>
using namespace std;
#define FREE (-1)
#define NOT_FREE (1)

int row[8];
int col[8];
int dpos[15];
int dneg[15];
#define N (8)

bool dfs( int r ) {
    if ( r == N ) return true;
    if ( row[r] != FREE ) {		// ?????§???????????????
	return dfs( r + 1 );
    }
    else {
	for ( int c = 0; c < N; c++ ) {
	    if ( col[c] == NOT_FREE ||
		 dpos[r + c] == NOT_FREE ||
		 dneg[r - c + N - 1] == NOT_FREE ) continue;
	    row[r] = c;
	    col[c] = NOT_FREE;
	    dpos[r + c] = NOT_FREE;
	    dneg[r - c + N - 1] = NOT_FREE;
	    if ( dfs( r + 1 ) ) return true;
	    row[r] = FREE;
	    col[c] = FREE;
	    dpos[r + c] = FREE;
	    dneg[r - c + N - 1] = FREE;
	}

	return false;
    }
}

int main()
{
    int k;
    scanf("%d", &k);
    for ( int i = 0; i < N; i++ ) row[i] = col[i] = FREE;
    for ( int i = 0; i < 2 * N - 1; i++ ) dpos[i] = dneg[i] = FREE;
    for ( int i = 0; i < k; i++ ) {
	int r, c;
	scanf("%d %d", &r, &c);
	row[r] = c;
	col[c] = NOT_FREE;
	dpos[r + c] = NOT_FREE;
	dneg[r - c + N - 1] = NOT_FREE;
    }

    dfs( 0 );

    for ( int r = 0; r < N; r++ ) {
	for ( int c = 0; c < N; c++ ) {
	    if ( row[r] == c ) {
		printf("Q");
	    }
	    else {
		printf(".");
	    }
	}
	printf("\n");
    }

    return 0;
}