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<stdio.h>
#define MAX 8
#define FREE -1
#define NOT_FREE 1

int Board[MAX][MAX], row[MAX], col[MAX], dpos[2*MAX-1], dneg[2*MAX-1];

void initialize() {
    for (int i=0; i<MAX; i++) row[i]=col[i]=FREE;
    for (int i=0; i<2*MAX-1; i++) dpos[i]=dneg[i]=FREE;
}

void printBoard() {
    for (int i=0; i<MAX; i++) {
        for (int j=0; j<MAX; j++) {
            if (row[i]!=j) printf(".");
            else printf("Q");
        }
        printf("\n");
    }
}

void recursive(int i) {
    if (i>=MAX) {
        printBoard();
        return;
    }
    else if (row[i]==FREE) {
        for (int j=0; j<MAX; j++) {
            if (row[i]!=FREE || col[j]!=FREE || dpos[i+j]!=FREE || dneg[i-j+MAX-1]!=FREE) continue;
            else {
                row[i]=j;
                col[j]=dpos[i+j]=dneg[i-j+MAX-1]=NOT_FREE;
                recursive(i+1);
                row[i]=FREE;
                col[j]=dpos[i+j]=dneg[i-j+MAX-1]=FREE;
            }
        }
    }
    else recursive(i+1);
}

int main() {
    int n, r, c;

    initialize();

    scanf("%d", &n);

    for (int i=0; i<n; i++) {
        scanf("%d %d", &r, &c);
        row[r] = c;
        col[c] = dpos[r+c] = dneg[r-c+MAX-1]=NOT_FREE;
    }

    recursive(0);
    return 0;
}
