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 <sstream>
#include <cstdio>
#include <stdlib.h>
#include <string>
#include <vector>
#include <algorithm>
#define N 8
using namespace std;
int row[N];
bool col[N],dpos[2*N-1],dneg[2*N-1];

void printBoard(){
  for(int i=0;i<N;i++){
    for(int j=0;j<N;j++){
      if(j==row[i])printf("Q");
      else printf(".");
    }
    printf("\n");
  }
}
void putqueen(int i){
  if(i == N){
    printBoard();
    return;
  }
  if (row[i]!=-1){
      putqueen(i+1);
  }
  else {
    for(int j=0;j<N;j++){
      if(col[j]||dpos[i+j]||dneg[i-j+N-1])continue;
      row[i]=j;
      col[j]=dpos[i+j]=dneg[i-j+N-1]= true;
      putqueen(i+1);
      row[i]=-1;
      col[j]=dpos[i+j]=dneg[i-j+N-1]= false;
    }
  }
}

int main(){
  int n,x,y;
  cin>>n;
  for(int i=0;i<N;i++)row[i]=-1;
  for(int i=0;i<n;i++){
    cin>>x>>y;
    row[x]=y;
    col[y]=dpos[x+y]=dneg[x-y+N-1]=true;
  }
  putqueen(0);

  return 0;
}

