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 <iostream>
using namespace std;

const int N = 8;
const int FREE = -1;
const int EXIST = 1;
int col[N], row[N], dpos[2*N], dneg[2*N];

void putQueen(int i){
    if (i == N){
        for (int j = 0; j < N; j++){
            for (int k = 0; k < N; k++){
                if (row[j] == k){
                    printf("Q");
                }else{
                    printf(".");
                }
            }
            printf("\n");
        }
        return;
    }
    //既にQueenが置いてあれば次の行へ進む
    if (row[i] != FREE) putQueen(i+1);
    else{
        for (int j = 0; j < N; j++){
            if (row[i] == FREE && col[j] == FREE && dpos[i+j] == FREE && dneg[i-j+N-1] == FREE){
                row[i] = j;
                col[j] = i;
                dpos[i+j] = dneg[i-j+N-1] = EXIST;
                putQueen(i+1);
                row[i] = col[j] = dpos[i+j] = dneg[i-j+N-1] = FREE;
            }
        }
    }
}

int main(){
    int n, i, j;
    cin >> n;
    for (i = 0; i < N; i++) row[i] = col[i] = FREE;
    for (i = 0; i < 2*N; i++) dpos[i] = dneg[i] = FREE;
    while(n--){
        cin >> i >> j;
        row[i] = j;
        col[j] = i;
        dpos[i+j] = dneg[i-j+N-1] = EXIST;
    }
    putQueen(0);

    return 0;
}

