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 FREE -1
#define NOT_FREE 1
int row[8],col[8],dir1[2*8-1],dir2[2*8-1];
bool x[8][8];

void init(){
  for(int i=0;i<8;i++){row[i]=FREE,col[i]=FREE;}
  for(int i=0;i<2*8-1;i++){dir1[i]=FREE,dir2[i]=FREE;}
}

void print(){
  for(int i=0;i<8;i++){
    for(int j=0;j<8;j++){
      if(x[i][j])if(row[i]!=j)return;
    }
  }
  for(int i=0;i<8;i++){
    for(int j=0;j<8;j++){
      if(row[i]==j) cout <<"Q";
      else cout << ".";
    }cout<<endl;
  }
}

void rec(int i){
  if(i==8){
    print();
    return;
  }

  for(int j=0;j<8;j++){
    if(col[j]==NOT_FREE || dir1[i+j] ==NOT_FREE || dir2[i-j+8-1]==NOT_FREE)continue;

    row[i]=j;
    col[j]=dir1[i+j]=dir2[i-j+8-1]=NOT_FREE;
    rec(i+1);
    row[i]=col[j]=dir1[i+j]=dir2[i-j+8-1]=FREE;
  }

}

int main(){
  init();
  for(int i=0;i<8;i++){
    for(int j=0;j<8;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;
  }

  rec(0);
  return 0;
}
