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<stdio.h>
#define N 8
#define FREE -1
#define NOT_FREE 1
void recursive(int);
void initialize();
void printBoard();
int row[N],col[N],dpos[2*N-1],dneg[2*N-1];

int X[N][N];
int main(){
  int r,c;
  initialize();

  for(int i=0;i<N;i++)
    for(int j=0;j<N;j++)
      X[i][j]=0;

  int k;
  scanf("%d",&k);
  for(int i=0;i<k;i++){
  scanf("%d%d",&r,&c);
  X[r][c]=1;
  }

recursive(0);
return 0;
}

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++){
      if(row[i]==j)
	printf("Q");
      else
      printf(".");}
    printf("\n");
  }
}

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;
  }
}


