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>
#include <vector>
#include <algorithm>
int k;
std::vector<int> r, c, que, used;
void solve(int i){
    if(i == 8){
        for(int j = 0; j < k; j++){
            if(que[r[j]] != c[j]) return;
        }
        for(int j = 0; j < 8; j++){
            for(int l = 0; l < 8; l++){
                if(j != l && std::abs(l - j) == std::abs(que[l] - que[j])) return;
            }
        }
        for(int j = 0; j < 8; j++){
            for(int l = 0; l < 8; l++){
                if(que[j] == l) std::cout << 'Q';
                else std::cout << '.';
                if(l == 7)std::cout << std::endl;
            }
        }
    }
    for(int j = 0; j < 8; j++){
        if(used[j] == 1) {
            que[i] = j;
            used[j] = 0;
            solve(i + 1);
            used[j] = 1;
        }
    }
}
int main(){
    std::cin >> k;
    r.resize(k);
    c.resize(k);
    used.resize(8, 1);
    que.resize(8);
    for(int i = 0; i < 8; i++) que[i] = i;
    for(int i = 0; i < k; i++) std::cin >> r[i] >> c[i];
    solve(0);
}
