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>
#include <cstring>
#include <algorithm>
using namespace std;
static const int FREE = -1;
static const int NOT_FREE = 1;
static const int N=8;

int row[N], col[N], dpos[2 * N - 1], dneg[2 * N -1];
bool X[N][N];

void init() {
    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;
    memset(X, 0, sizeof (X));
}

void printBoard() {
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
            if (X[i][j] && row[i] != j) return;
        }
    }

    for (int i = 0; i < N; i++) {
        for (int j = 0; j  < N; j++)
            (row[i] == j) ? printf("Q") : printf(".");
        printf("\n");
    }
}

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

int main()
{
    int k, r, c;
    init();
    scanf("%d", &k);
    for (int i = 0; i < k; i++) {
        scanf("%d %d", &r, &c);
        X[r][c] = true;
    }
    solve(0);
    return 0;
}

