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>
using namespace std;
vector<int> cond(8,-1), ans(8);
bool valid(int pos){
    bool ret = true;
    for(int i = 0; i < pos; i++){
        ret &= ans[i] != ans[pos];
        ret &= abs(ans[pos]-ans[i]) != pos-i;
    }
    return ret;
}
void dfs(int pos){
    if(pos == 8){
        for(int i = 0; i < 8; i++){
            if(cond[i] != -1 && cond[i] != ans[i])  return;
        }
        // found!
        for(int i = 0; i < 8; i++){
            for(int j = 0; j < 8; j++){
                if(ans[i] == j)    cout << "Q";
                else               cout << ".";
            }
            cout << endl;
        }
        exit(0);
    }
    for(int j = 0; j < 8; j++){
        ans[pos] = j;
        if(!valid(pos))  continue;
        dfs(pos+1);
    }
}
int main(){
    int n;cin >> n;
    while(n-- > 0){
        int i, j;
        cin >> i >> j;
        cond[i] = j;
    }
    dfs(0);
}
