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<cassert>
using namespace std;

#define N 8
#define FREE -1
#define NOT_FREE 1

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

bool X[N][N];

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

void printBoard(){
    for(int i = 0;i<N;i++){
        for(int j = 0;j<N;j++){\
        if(X[i][j]){
            if(row[i] != j) return;
        }
    }
}
for(int i = 0;i<N;i++){
    for(int j = 0;j<N;j++){
        cout << ( (row[i] == j)? "Q" : ".");
    }
    cout << endl;
  }
}

void recursive(int i){
    if(i == N){
        printBoard(); return;
}

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

int main(){
    initialize();
    
    for(int i = 0;i<N;i++)
    for(int j = 0;j<N;j++) X[i][j] = false;
    
    int k; cin >> k;
    for(int i = 0;i<k;i++){
        int r, c; cin >> r >> c;
        X[r][c] = true;
    }
    
    recursive(0);
    
    return 0;
}
