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 Notfree 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 printB(){
  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){
    printB();
    return;
  }
  for(int j=0;j<N;j++){
    if(Notfree==col[j] || Notfree==dpos[i+j] || Notfree==dneg[i-j+N-1])continue;

    row[i]=j;
    col[j]=dpos[i+j]=dneg[i-j+N-1]=Notfree;
    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<N;i++){
    int r,c;
    cin>>r>>c;
    x[r][c]=true;
  }
  recursive(0);

  return 0;
}

