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 N_FREE 1


int low[N],co[N],dps[2*N-1],dng[2*N-1];
bool Y[N][N];

void initialize(){
  int i;

  for(i=0;i<N;i++){
    low[i] = FREE;
    co[i] = FREE;
  }

  for(i=0;i<2*N-1;i++){
    dps[i] = FREE;
    dng[i] = FREE;
  }
}

void print(){
  int i,j;

  for(i=0;i<N;i++){
    for(j=0;j<N;j++){
      if(Y[i][j]){
        if(low[i] != j) return;
      }
    }
  }

  for(i=0;i<N;i++){
    for(j=0;j<N;j++){
        cout << ((low[i] == j) ? "Q" : ".");

    }
cout << endl;
  }

}

void recursive(int l){
int j;

  if(l==N){
    print(); return;
  }
for(j=0;j<N;j++){
  if(N_FREE == co[j] || N_FREE == dps[l+j] || N_FREE == dng[l-j+N-1]) continue;
  low[l]=j;
  co[j]=dps[l+j]=dng[l-j+N-1] = N_FREE;
  recursive(l+1);
  low[l]=co[j]=dps[l+j]=dng[l-j+N-1] = FREE;
}

}



int main(){
  int i,j,n,q,p;

  initialize();

  for(i=0;i<N;i++){
    for(j=0;j<N;j++){
      Y[i][j]= false;
    }
  }
  cin >> n;

  for(i=0;i<n;i++){
  cin >> p >> q;
    Y[p][q] = true;
  }

  recursive(0);


  return 0;
}

